AP Physics B & C - Multiple Choice Practice Questions on Work, Energy and Power

“The best thinking has been done in solitude. The worst has been done in turmoil.”

– Thomas A. Edison

Today we will discuss a few multiple choice practice questions involving work, energy and power. Questions in this section were discussed earlier on this site. You can access them by clicking on the label ‘work energy power’ or by trying a search for the required word using the search box provided on this page.

(1) A sphere A of mass m moves with speed v and kinetic energy E along the positive x-direction and collides inelastically with another identical sphere B at rest. After the collision sphere A moves with kinetic energy E/3 along the positive y-direction. What is the speed of sphere B after the collision?

(a) v/2

(b) v(√3)/2

(c) v(√5)/2

(d) 2v/√3

(e) 4v/√3

We have E = ½ mv²

Since the kinetic energy is directly proportional to the square of speed, the speed of A after the collision is v/√3.

Momentum is conserved in all collisions (elastic as well as inelastic). The momentum of sphere A before the collision is mv and is directed along the positive x-direction. This is the total momentum of the system since the initial momentum of sphere B is zero.

The momentum mv is represented by the vector OP₁ in the adjoining figure. The momentum of sphere A after the collision is mv/√3 and is directed along the positive y-direction. This is represented by the vector OP₂. The momentum of sphere B after the collision must be given by the vector OP₃ since the parallelogram OP₂P₁P₃ is drawn such that its diagonal OP₁ represents the final momentum, mv of the system

Evidently │OP₃‌│² = │OP₁│² + │OP₂│²

Or, │OP₃‌│² = (mv)² + (mv/√3)²

But │OP₃‌│ = mv_f where v_f is the speed of sphere B after the collision.

Thus we have

(mv_f)² = (mv)² + (mv/√3)²

This gives v_f = √(v² + v²/3) = 2v/√3

(2) A particle moves in a plane under the action of a force of constant magnitude. If the direction of the force is always at right angles to the direction of motion of the particle, it follows that

(a) the momentum of the particle is constant

(b) the velocity of the particle is constant

(c) the kinetic energy of the particle is constant

(d) the acceleration of the particle is constant

(e) the particle moves along a straight line with increasing speed

A force of constant magnitude acting always at right angles to to the direction of motion of the particle will supply centripetal force required for uniform circular motion. In uniform circular motion the speed of the particle is constant.

[The velocity is not constant since the direction changes continuously. The momentum and the acceleration also are not constant for the same reason].

Since the speed is constant, the kinetic energy of the particle is constant [Option (c)].

(3) A ball of mass 0.5 kg is thrown vertically downwards from the edge of a building of height 65 m. The initial speed of the ball is 20 ms^–1. If the ball strikes the ground with a speed of 40 m/s, the energy lost by the ball because of air resistance is (gravitational acceleration, g = 10 ms^–2)

(a) 10 J

(b) 25 J

(c) 30 J

(d) 40 J

(e) 45 J

The initial energy of the ball is E_i = mgH + ½ mv_i ²

The first term is the gravitational potential energy of the ball of mass m at height H and the second term is the initial kinetic energy of the ball having initial speed v_i. Substituting proper values, we have,

E_i = 0.5×10×65 + ½ ×0.5×20² = 325 + 100 = 425 J.

The final energy of the ball (at the ground level) having velocity v_f is E_f = ½ mv_f²

Substituting proper values, we have,

E_f = ½×0.5×40² = 400 J.

Therefore, the energy lost by the ball because of air resistance is E_i – E_f = 425 J – 400 J = 25 J.

(4) Two blocks A and B of masses m and 3m respectively (Fig.) travel in opposite directions with the same speed v along a smooth surface. They collide head-on and stick together. Mechanical energy lost during the collision is

(a) ½ mv²

(b) mv²

(c) (³/₂) mv²

(d) 2 mv²

(e) 3 mv²

Total momentum of the system before collision is 3mv – mv = 2 mv

[We have taken the leftward momentum as positive and that’s why the total momentum is positive. If you take the leftward momentum as negative, the total momentum will be negative. But this won’t affect your final answer and the energy lost during the collision will be positive].

Since the momentum of the system is conserved, the total momentum after the collision will be 2 mv itself. If the common velocity of A and B (which stick together) after collision is v_f, we have,

(m + 3m)v_f = 2 mv

Therefore v_f = v/2

The initial kinetic energy of the system is ½ mv² + ½ (3m)v² = 2 mv²

Final kinetic energy of the system is ½ (4m)v_f² = ½ (4m)(v/2)² = ½ mv²

Therefore, the loss of kinetic energy during the collision = 2 mv² – ½ mv² = (³/₂) mv²

(5) An electric motor creates a tension of 1000 N in a hoist cable and reels it at the rate of 2.5 ms^–1. The power output of the motor is

(a) 250W

(b) 400 W

(c) 2.5 kW

(d) 2500 kW

(e) 400 kW

Since the point of application of the force of 1000 N moves through 2.5 m per second, the work done per second (which is the power output) is 1000×2.5 watt = 2500 watt = 2.5 kW

Answer to AP Physics C Free Response Practice Question on Rotation

In the post dated 30^th September 2009 A free response question (for practice) on rotational motion for AP Physics C aspirants was given. As promised I give below a model answer for the same. The question also is given:

A solid cylinder of mass M and radius R has a light inextensible string wound round it. The free end of the string is tied to a rigid support (Fig.) and the cylinder is released from its state of rest.

(a) Determine the acceleration ‘a’ of the cylinder as it moves down, unwinding the string.
The cylinder itself is now suspended from the support so that it is free to rotate about its axle. The friction at the axle of the cylinder is negligible. A mass M₁ is now attached to the free end of the string wound round the cylinder as shown in the adjoining figure and the system is released from rest at time t = 0. Now answer the following questions (b), (c) and (d):

(b) Determine the acceleration a₁ of the mass M₁ in terms of the given parameters.

(c) In terms of the given parameters obtain the angular acceleration of the cylinder when the mass M₁ moves down, unwinding the string.

(d) Determine the kinetic energy of the cylinder at time t.

(a) The weight Mg of the cylinder tries to drive it downwards while the tension T in the sring tries to oppose the downward motion (fig.). The net force on the cylinder is Mg – T. If the downward acceleration is a we have

Mg – T = Ma from which T = M (g – a)

The torque produced by the tension T rotates the cylinder and produces an angular acceleration α (let us say) so that we have

TR = Iα where I is the moment of inertia (MR²/2) of the cylinder about its axis (axle). Therefore, TR = (MR²/2)α

But α = a/R so that TR = (MR²/2)(a/R)

Since T = M (g – a), the above equation becomes

M (g – a)R = (MR²/2)(a/R) from which (g – a) = a/2 so that a = 2g/3

(b) The force driving the mass M₁ downwards is its weight M₁g. The tension T₁ in the sring tries to oppose the downward motion (fig.). The downward acceleration a₁ of the mass M₁ is therefore given by M₁g – T₁ = M₁a₁ from which

T₁ = M₁(g – a₁)

Since the torque produced by the tension T₁ is T₁R and this is responsible for the rotation of the cylinder, we have

T₁R = Iα₁ where α₁ is the angular acceleration of the cylinder.

Substituting for T₁ and the moment of inertia I, we have

M₁(g – a₁)R = (MR²/2)α₁

Substituting for α₁ = a₁/R, the above equation becomes

M₁(g – a₁)R = (MR²/2)(a₁/R)

Or, 2M₁(g – a₁) = Ma₁, from which a₁ = 2M₁g/(2M₁+M)

(c) The angular acceleration α₁ of the cylinder when the mass M₁ moves down, unwinding the string is given by

α₁ = a₁/R = 2M₁g/ [(2M₁+M)R]

(d) The kinetic energy of the cylinder at time t is due to its rotational motion and is equal to ½ Iω² where ω is its angular velocity at time t.

But ω is related to the angular acceleration α₁ of the cylinder as

α₁ = dω/dt

Therefore, dω/dt = 2M₁g/ [(2M₁+M)R]

Integrating, ω =2M₁gt / [(2M₁+M)R] + C where C is the constant of integration which can be found out from the initial condition.

Initially (when t = 0) the cylinder is at rest and hence ω = 0. Substituting in the above equation we have

0 = 0 + C. Therefore C = 0 and the expression for ω becomes

ω =2M₁gt / [(2M₁+M)R]

[Since the angular acceleration is constant, you can easily obtain ω from the equation,

ω = ω₀ + αt where ω₀ = 0 and α = α₁ = 2M₁g/{(2M₁+M)R}]

Therefore, kinetic energy of cylinder = ½ Iω² = ½ ×(MR²/2) ×4M₁²g²t² / [(2M₁+M)R]².

This yields the value MM₁²g²t² / (2M₁+M)².

* * * * * * * * * * * * * * * * * *

Suppose you were asked to determine the kinetic energy of the mass M₁ at time t. You will then proceed as follows:

At time t the mass M₁ has kinetic energy equal to ½ M₁v² where v is the velocity acquired by it in time t while moving down with acceleration a₁. Since it started from rest we have

v = v₀ + a₁t = 0 + [2M₁g/(2M₁+M)]t

Therefore, kinetic energy of mass M₁ = ½ M₁v² = ½ ×M₁[2M₁gt /(2M₁+M)]²

This yields the value 2M₁³g²t² /(2M₁+M)²

AP Physics C – A Free Response Practice Question on Rotation

Essential points and some practice questions on rotational motion have been discussed on this site earlier. You can access all posts related to rotational motion on this site by clicking on the label ‘rotation’ below this post.

Today we will discuss a free response question (for practice) in this section. This

question is meant for AP Physics C aspirants:

A solid cylinder of mass M and radius R has a light inextensible string wound round it. The free end of the string is tied to a rigid support (Fig.) and the cylinder is released from its state of rest.

(a) Determine the acceleration ‘a’ of the cylinder as it moves down, unwinding the string.

The cylinder itself is now suspended from the support so that it is free to rotate about its axle. The friction at the axle of the cylinder is negligible. A mass M₁ is now attached to the free end of the string wound round the cylinder as shown in the adjoining figure and the system is released from rest at time t = 0. Now answer the following questions (b), (c) and (d):

(b) Determine the acceleration a₁ of the mass M₁ in terms of the given parameters.

(c) In terms of the given parameters obtain the angular acceleration of the cylinder when the mass M₁ moves down, unwinding the string.

(d) Determine the kinetic energy of the cylinder at time t.

Try to answer this question which carries 15 points. You have 15 minutes at your disposal. I’ll be back with a model answer for you shortly.

Meanwhile find some useful multiple choice questions with solution in this section at physicsplus.