A.) 1.0 s
B.) 2.0 s
C.) 4.0 s
D.) 5.0 s
E.) 10.0 s
Answer: B
Explanation: The time taken for the ball to reach its maximum height can be calculated using the formula t = v₀/g, where v₀ is the initial velocity and g is the acceleration due to gravity. Thus, t = 20 m/s / 9.8 m/s² = 2.04 s ≈ 2.0 s.
A.) 2 m/s²
B.) 4 m/s²
C.) 6 m/s²
D.) 8 m/s²
E.) 10 m/s²
Answer: E
Explanation: he magnitude of the acceleration of the car during braking can be calculated using the formula a = (v² – v₀²)/2x, where v₀ is the initial velocity, v is the final velocity (which is zero in this case), and x is the displacement. Thus, a = (0 – 20 m/s)² / (2 * 50 m) = 10 m/s².
A.) 10 m
B.) 20 m
C.) 30 m
D.) 40 m
E.) 50 m
Answer: D
Explanation: The maximum height reached by the projectile can be calculated using the formula h = (v₀sinθ)²/2g, where v₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. Thus, h = (20 m/s * sin 30°)² / (2 * 9.8 m/s²) = 40.8 m ≈ 40 m.
A.) 2gx/k
B.) (2gx/k)½
C.) gx/k
D.) (gx/k)½
E.) (2gx/k)¼
Answer: B
Explanation: The maximum potential energy of the block is equal to the maximum kinetic energy of the block as it passes through the equilibrium position. The maximum potential energy is equal to (1/2)kx2, and the maximum kinetic energy is equal to (1/2)Mv2, where v is the maximum speed of the block. Equating these two expressions and solving for v gives v = (2gx/k)½.
A.) v2/2g
B.) v/2g
C.) v2/g
D.) v/g
E.) 2v/g
Answer: A
Explanation: At the maximum height, the vertical component of the ball’s velocity is zero. Therefore, using the equation of motion v2 = u2 + 2gs, where u is the initial velocity, v is the final velocity (which is zero), g is the acceleration due to gravity, and s is the displacement, we get s = v2/2g.
A.) (1/2)kA2
B.) (1/2)MA2
C.) (1/2)kA2 + (1/2)Mv2
D.) (1/2)kA2 + (1/2)MA2
E.) (1/2)kA2 + (1/2)Mv2 + MgA
Answer: D
Explanation: The maximum potential energy of the system is equal to the maximum potential energy stored in the spring at maximum displacement plus the maximum potential energy stored in the block at maximum displacement. The maximum potential energy stored in the spring is (1/2)kA2, and the maximum potential energy stored in the block is (1/2)MA2. Therefore, the maximum potential energy of the system is (1/2)kA2 + (1/2)MA2.
A.) (k/M)½
B.) (M/k)½
C.) (g/k)½
D.) (k/Mg)½
E.) (Mg/k)½
Answer: B
Explanation: The natural frequency of the system is equal to (1/2π)(k/M)½, where k is the spring constant and M is the mass of the block. Therefore, the natural frequency of the system is (M/k)½.
A.) Fd
B.) Fd/m
C.) 1/2 Fd
D.) 1/2 Fd/m
E.) None of the above
Answer: A
Explanation: The work done by a force is equal to the force multiplied by the displacement in the direction of the forcE.) Since the surface is frictionless, there is no work done by friction, and the work done by the force F is simply Fd.
A.) 50 W
B.) 100 W
C.) 150 W
D.) 200 W
E.) 250 W
Answer: B
Explanation: Since the crate is being lifted at a constant speed, the net work done on the crate is zero. Therefore, the power delivered by the tension in the rope is equal to the weight of the crate multiplied by the lifting speed The weight of the crate is 50 kg x 9.8 m/s² = 490 N, and the lifting speed is 2 m/s. Thus, the power delivered by the tension in the rope is 490 N x 2 m/s = 980 W or 100 W.
A.) Mgh
B.) Fh
C.) F/mgh
D.) Fmg/h
E.) None of the above
Answer: B
Explanation: The work done by a constant force is equal to the force multiplied by the displacement in the direction of the force. In this case, the force is applied in the upward direction and the displacement is also upward, so the work done by the force F is simply Fh.
A.) 5 m
B.) 10 m
C.) 15 m
D.) 20 m
E.) 25 m
Answer: C
Explanation: The maximum height reached by the object can be found using the formula h = v²/2g, where v is the initial velocity of the object and g is the acceleration due to gravity. Therefore, h = (10 m/s)² / (2 x 9.8 m/s²) = 5.1 m. However, the object is thrown vertically upward, so it also experiences a deceleration due to gravity, which reduces its upward velocity to zero at the maximum height.
A.) 8 m/s
B.) 10 m/s
C.) 15 m/s
D.) 20 m/s
E.) 25 m/s
Answer: B
Explanation: Using conservation of momentum, we have (10 kg)(5 m/s) = (2 kg)v, where v is the final velocity of the 2 kg object. Solving for v gives v = 25/2 m/s. However, since the 10 kg object comes to rest after the collision, the final momentum of the system must be zero. Therefore, the final velocity of the 2 kg object is also zero, so the correct answer is B.
A.) -10 m/s
B.) -5 m/s
C.) 0 m/s
D.) 5 m/s
E.) 10 m/s
Answer: A
Explanation: Using conservation of momentum, we have (2 kg)(10 m/s) = (3 kg)v, where v is the final velocity of the two objects. Solving for v gives v = 6.67 m/s. Since the 2 kg object is moving at 5 m/s after the collision, the change in velocity of the 1 kg object is -15 m/s, which means its final velocity is -10 m/s. Therefore, the correct answer is A.
A.) -4 m/s
B.) -3 m/s
C.) -2 m/s
D.) -1 m/s
E.) 0 m/s
Answer: D
Explanation: Using conservation of momentum, we have (1 kg)(5 m/s) + (2 kg)(-3 m/s) = (1 kg)(2 m/s) + (2 kg)v, where v is the final velocity of the two objects. Solving for v gives v = -1 m/s. Therefore, the correct answer is D.
A.) 2 m/s
B.) 4 m/s
C.) 6 m/s
D.) 8 m/s
E.) 12 m/s
Answer: A
Explanation: Since the two objects stick together after the collision, the momentum of the system is conserved. Using conservation of momentum, we have (2m)(4 m/s) + (0) = (2m)v, where v is the velocity of the combined object after the collision. Solving for v gives v = 2 m/s. Therefore, the correct answer is A.
A.) v = sqrt(2gh)
B.) v = sqrt(gh)
C.) v = (5/7) sqrt(2gh)
D.) v = (3/5) sqrt(2gh)
E.) v = (7/5) sqrt(2gh)
Answer: D
Explanation: The total mechanical energy of the disk at the top of the incline is given by E = Mgh, where h is the height of the incline. At the bottom of the incline, all of this energy is kinetic energy, which is equal to the sum of the translational kinetic energy and the rotational kinetic energy of the disk. Using the conservation of energy, we can find the speed of the center of mass of the disk at the bottom of the incline to be v = (3/5) sqrt(2gh).
A.) It is doubled
B.) It is halved
C.) It is quadrupled
D.) It is unchanged
E.) It is multiplied by a factor of 8
Answer: C
Explanation: The moment of inertia of a disk about its center is given by I = (1/2) MR2. If the radius of the disk is doubled, the moment of inertia becomes I’ = (1/2) M(2R)2 = 2MR2. If the angular speed is halved, the new angular velocity becomes ω’ = ω/2. The new moment of inertia is therefore I’ω’ = 2MR2(ω/2) = MR2ω, which is four times the original moment of inertia.
A.) F = Mg/2
B.) F = Mg/3
C.) F = Mg/4
D.) F = Mg/5
E.) F = Mg/6
Answer: C
Explanation: The force required to make the disc roll without slipping is equal to the frictional force between the disc and the ground. The maximum static frictional force is given by fs = μs N, where μs is the coefficient of static friction and N is the normal force on the disc. At the minimum force required to make the disc roll without slipping, the frictional force is equal to the force applied tangentially to the disc, i.e., F = fs. The normal force N is equal to the weight of the disc, i.e., N = Mg. Therefore, we have F = μs Mg. For a uniform circular disc rolling without slipping, the coefficient of static friction is μs = 1/4. Hence, the minimum force required to make the disc roll without slipping is F = (1/4) Mg.
A.) 0.125 s
B.) 0.25 s
C.) 0.5 s
D.) 1 s
E.) 2 s
Answer: E
Explanation: The period of a spring-mass system is given by T = 2π√(m/k), where m is the mass and k is the spring constant. Since the mass is increased by a factor of 4, the period will increase by a factor of 2, since T is inversely proportional to the square root of m.
A.) 0.5 Hz
B.) 1 Hz
C.) 2 Hz
D.) 4 Hz
E.) 8 Hz
Answer: B
Explanation: The frequency of the system is given by f = 1/(2π)√(k/m), where m is the reduced mass of the system, which is given by m = (m1m2)/(m1+m2). Substituting the given values, we get the reduced mass to be 1.2 kg. Substituting this and the given values of k, we get the frequency to be 1 Hz.
A.) 0.2 m/s²
B.) 2 m/s²
C.) 20 m/s²
D.) 200 m/s²
E.) 200
Answer: B
Explanation: The maximum acceleration of the block is equal to the maximum force acting on the block divided by its mass. The maximum force is equal to the maximum restoring force of the spring, which is given by F = kx, where x is the maximum displacement of the block. Substituting the given values, we get F = 20 N. Dividing by the mass of the block, we get the maximum acceleration to be 2 m/s².
A.) 0.1 J
B.) 0.4 J
C.) 0.8 J
D.) 1 J
E.) 2 J
Answer: B
Explanation: The maximum potential energy of the block is equal to the maximum potential energy stored in the spring, which is given by (1/2)kx², where x is the amplitude of the oscillation. Substituting the given values, we get the maximum potential energy to be 0.4 J.
A.) 0.25 rad/s
B.) 0.5 rad/s
C.) 1 rad/s
D.) 2 rad/s
E.) 4 rad/s
Answer: C
Explanation: The angular frequency of a simple pendulum is given by ω = √(g/L), where g is the acceleration due to gravity and L is the length of the pendulum. Since the pendulum is displaced by an angle of 30°, the effective length of the pendulum is L cos(30°) = 0.866 m. Substituting this and the given value of g, we get the angular frequency to be 1 rad/s.
A.) The planet’s distance from the sun cubed
B.) The planet’s mass
C.) The gravitational constant
D.) The planet’s distance from the sun
E.) The planet’s mass squared
Answer: A
Explanation: Kepler’s Third Law states that the square of the period of revolution of a planet around the sun is proportional to the cube of its semi-major axis.
A.) continue in its circular orbit at a constant speed
B.) move outwards to a higher altitude
C.) move inwards to a lower altitude
D.) move away from the Earth in a straight line
E.) move towards the Earth and eventually crash
Answer: D
Explanation: In the absence of a gravitational force, an object in motion will continue to move in a straight line with constant speed.
A.) The distance between them
B.) The product of their masses
C.) The inverse square of the distance between them
D.) Both A and B
E.) Both B and C
Answer: E
Explanation: The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
A.) 6.4 m/s2
B.) 9.8 m/s2
C.) 10.7 m/s2
D.) 12.5 m/s2
E.) 15.3 m/s2
Answer: B
Explanation: The acceleration due to gravity at the surface of a planet is given by g = GM/r2, where G is the gravitational constant, M is the mass of the planet, and r is its radius.
A.) 5 Nm
B.) 10 Nm
C.) 15 Nm
D.) 20 Nm
E.) 25 Nm
Answer: B
Explanation: The torque produced by a force applied tangentially to a wheel is given by the product of the force and the radius of the wheel. Thus, the torque in this case is (10 N) x (0.5 m) = 5 Nm.
A.) √(2gh)
B.) √(gh)
C.) √(4gh/3)
D.) √(3gh/4)
E.) √(5gh/7)
Answer: A
Explanation: The potential energy of the cylinder at the top of the incline is given by Mgh, and the kinetic energy of the cylinder at the bottom of the incline is given by (1/2)MV² + (1/2)Iω², where I is the moment of inertia of the cylinder and ω is its angular velocity. Since the cylinder is rolling without slipping, V = Rω. Using the parallel axis theorem, the moment of inertia of the cylinder about an axis through its center of mass is (1/2)MR². Solving for V, we get V = √(2gh).
A.) √(2L/g)
B.) √(3L/2g)
C.) √(4L/3g)
D.) √(5L/4g)
E.) √(6L/5g)
Answer: C
Explanation: The period of oscillation of a simple pendulum is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. For a uniform rod pivoted at one end, the effective length of the pendulum is (1/3)L. Thus, the period of oscillation is T = 2π√((1/3)L/g) = √(4L/3g).
A.) 2π√(M/k)
B.) π√(M/k)
C.) √(2πM/k)
D.) √(2M/πk)
E.) √(M/2πk)
Answer: A
Explanation: The period of oscillation of a mass-spring system is given by T = 2π√(m/k), where m is the effective mass of the system, which is equal to the mass of the block in this case. Thus, the period of oscillation is T = 2π√(M/k).
A.) 2 rad/s²
B.) 4 rad/s²
C.) 6 rad/s²
D.) 8 rad/s²
E.) 10 rad/s²
Answer: C
Explanation: The moment of inertia of a uniform disk is (1/2)MR². Thus, the angular acceleration produced by a torque is given by α = τ / I = (10 Nm) / [(1/2)(2 kg)(0.5 m)²] = 6 rad/s².
A.) Determine the normal force, gravitational force, and force of tension in the incline.
Answer:
The normal force on the block is equal to the component of the gravitational force perpendicular to the incline. Therefore, we have FN = mg cos(30) = 34.6 N.
The gravitational force acting on the block is equal to the weight of the block. Therefore, we have Fg = mg sin(30) = 19.6 N.
The force of tension in the incline is equal to the component of the force applied to the block parallel to the incline. Therefore, we have FT = Fapplied x sin(45) = 14.1 N.
B.) Determine the acceleration of the block and the force parallel to the incline acting on the block.
Answer:
The net force acting on the block parallel to the incline is equal to the force of tension minus the component of the gravitational force parallel to the incline.
Therefore, we have Fnet = FT – Fg x sin(30) = 4.5 N.
The acceleration of the block is equal to the net force divided by the mass of the block.
Therefore, we have a = Fnet /m = 1.125 m/s2.
C.) Determine the velocity of the block when it reaches the bottom of the incline.
Answer:
The velocity of the block when it reaches the bottom of the incline can be determined using kinematic equations. Since the block starts from rest, we have vf2 = vi2 + 2ad, where vf = 0, a = 1.125 m/s2, and d = l/sin(30) = 8 m.
Therefore, we have vf = √(2×1.125) = 4.24 m/s
D.) The block rebounds from the incline with a speed of 1.5 m/s. Determine the time of contact between the block and the incline during the collision.
Answer:
The coefficient of restitution between the block and the incline is given by e = vf/vi, where vi is the velocity of the block before the collision.
Therefore, we have vi = vf/e = 10.56 m/s.
The time of contact between the block and the incline during the collision can be determined using the formula t = 2vi*sin(45)/g, where g is the acceleration due to gravity.
Therefore, we have t = 0.338 s.
E.) Determine the impulse experienced by the block during the collision.
Answer:
The impulse experienced by the block during the collision can be determined using the formula J = Favg(t), where Favg is the average force during the collision. Since the collision is elastic, the average force is equal to the net force during the collision.
Therefore, we have Favg = Fnet = 4.5 N, and J = Favg(t) = 1.52 Ns.
F.) Determine the height the block reaches after bouncing off the incline.
Answer:
The height the block reaches after bouncing off the incline can be determined using energy conservation. Since the collision is elastic, the total energy of the block-incline system is conserved. Therefore, we have mgh = 1/2mvf2, where h is the height the block reaches after bouncing off the incline.
Solving for h, we have h = vf2/(2g) = 3.42 m.
A.) Draw a free body diagram for the forces acting on the hockey puck and label all of them.
Answer:
The forces acting on the hockey puck are: the force of gravity mg acting downward, the normal force N perpendicular to the surface, the force of friction f parallel to the surface and opposite in direction to the motion, and the applied force F at an angle θ above the horizontal.
B.) Compute the normal force acting on the hockey puck in terms of m, θ, and g.
Answer:
First, we can write the Friction force F in terms of its horizontal and vertical components: F sinθ and F cosθ:
Since the hockey puck is not moving upwards or downwards, which means that there is no net vertical force, the normal force N and the force of gravity mg is equal to each other. Therefore we have this expression:
N + F sinθ = mg → N = mg – F sinθ
C.) Compute the acceleration of the hockey puck in terms of m, F, θ, μ, and g.
Answer:
We know that the hockey puck has a net horizontal movement. Therefore, it can be depicted by this expression:
F cosθ – Friction force f
This can be written as F cosθ – μ(N) since frictional force is calculated by multiplying the normal force with the coefficient of kinetic friction.
Since the expression needs to be written in terms of m, F, θ, μ, and mg, we can see from part (b) that the normal force N is written as mg – F sinθ
Therefore, substituting N, we have Net force = F cosθ – μ(mg – F sinθ)
To find the acceleration, the net force needs to be divided by the mass. Therefore we have: (F cosθ – μ(mg – F sinθ)) / m
D.) Assume that the magnitude of the applied force is fixed but that the angle may be varied. For what value of θ would the resulting horizontal acceleration of the hockey puck be maximized?
Answer:
The expression for the net force needs to be maximized.
Remembering what we derived: F cosθ – μ(mg – F sinθ).
When expanded, we have F cosθ – μmg + μF sinθ
Rearranging it, we have F (cosθ + μsinθ) – μ(mg )
Since F (the applied force), μ (coefficient of kinetic friction), m (mass) and mg (force of gravity) are all constant values, only the expression cosθ + μsinθ can be maximized.
To maximize it, it means that the derivative of the expression needs to be differentiated and equated to zero. To find the derivative of f(θ) = cosθ + μsinθ = 0, we can use the chain rule and the derivative of sine and cosine functions:
f'(θ) = -sinθ + μcosθ = 0
μcosθ = sinθ → μ = sinθ/cosθ → μ = tanθ
θ = tan-1μ
This shows that f”(θ) = -cosθ – μsinθ will always be negative for the θ interval of 0 ≤ θ ≤ ½ π. Further substituting θ = tan-1μ, we have:
f(tan-1μ) = cos(tan-1μ) + μsin(tan-1μ)
1/√(1+μ2) + μ(1/√(1+μ2) = √(1+μ2)
This expression is greater than f(0) = 1 or f (½ π) = μ where 1 and μ are the values of f at the endpoints of the interval
A.) What event occurred to the car at t = 2 s?
Answer:
The velocity of the car begins to decrease at t = 1 s, which is indicated by a change in acceleration from positive to negative. This change in acceleration is reflected in the slope of the given velocity-versus-time graph.
B.) How does the car’s average velocity between t = 0 and t = 2 s compare to the average velocity between t = 2 s and t = 8 s?
Answer:
Average velocity between t = 0 s and t = 2 s is (vt=0 + vt=2 ) / 2 = (0+20)/2 = 10 𝑚/s
Average velocity between t = 2 s and t = 8 s is (vt=2 + vt=8) / 2 = (20+0)/2 = 10 𝑚/s
Therefore, it can be seen that the average velocities are the same.
C.) Determine the displacement of the car between t = 0 and t = 8s
Answer:
To determine the displacement of an object as a function of time, we can use the fact that the displacement is equal to the area under the velocity-time curve, with areas above the time axis being considered positive and those below being negative.
In the given scenario, the displacement between t = 0 and t = 8 s is calculated by finding the area of the triangle that is formed by the velocity-time graph and the time axis, where the base of the triangle corresponds to the time interval from t = 0 to t = 8 s.
Area of the positive triangle space in the graph = ½ x b x h where b = 8 and h = 20. Therefore, ½ x 8 x 20 = 80 m
Area of the negative triangle space in the graph = -½ x b x h where b = 3 and h = 10. Therefore, -½ x 3 x (10) = -15 m
Hence, the total displacement is the addition of both distances: 80 + (-15) = 65 m
D.) Plot the car’s acceleration during this time interval as a function of time.
Answer:
The acceleration is the slope of a v vs t graph. Therefore, the graph segment:
t = 0 s to t = 2 s is a = (20-0)/(2-0) = 10 m/s2
t = 2 s to t = 11 s is a = (-10-20)/(11-2) = -3.3 m/s2
Therefore, drawing the car’s acceleration in a a vs t graph:
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