The essential points you have to remember to answer multiple choice questions on work, energy and power were discussed in the post dated 14

^{th}November 2008. Today we will discuss some multiple choice questions in this section.The following questions are meant

*for***aspirants***AP Physics B & C***(1)**A small doll suspended inside a truck using a string of length

*ℓ*, undergoes an angular displacement

*θ*when the truck accidentally collides against a wall and stops. If the truck was moving with uniform velocity towards the wall before the collision, the magnitude of the velocity of the truck was

(a) [2

*g**ℓ*(1– sin*θ*)]^{1/2}(b) 2

*g**ℓ*(1– sin*θ*)](c) [2

*g**ℓ*(1– cos*θ*)]^{1/2}(d) 2

*g**ℓ*(1– cos*θ*)The kinetic energy of the doll (which serves as the bob of a simple pendulum) is converted into gravitational potential energy when the truck stops. In other words, the bob moves up doing work against the gravitational force, losing its kinetic energy and gaining an equivalent potential energy.

If the maximum height attained by the doll is

*h*(fig.), we have ½

*mv*^{2}=*mgh*where*m*is the mas of the doll,*v*is*the velocity of the truck and**g*is the acceleration due to gravity. Therefore,*v*= (2*gh*)^{1/2}^{ }.But

*h =**ℓ*–*ℓ*cos*θ**=**ℓ*(1– cos*θ*).Substituting for

*h*we have*v*= [2*g**ℓ*(1– cos*θ*)]^{1/2}**(2)**The kinetic energy of a moving particle at time

*t*is found to be directly proportional to

*t*

^{2}. The force acting on the particle is directly proportional to

(a)

*t*^{2}(b)

*t*^{3/2}(c)

*t*(d)

*t*^{1/2}(e)

*t*^{0}since the kinetic energy is directly proportional to

*t*^{2}, we can write ½

*mv*^{2}=*C t*^{2}where*m*and*v*are the mass and velocity of the particle and*C*is the constant of proportionality.Therefore,

*v*= [√(2*C/m*)]*t*The velocity is thus directly proportional to the time

*t*which means that the particle is in uniformly accelerated motion. The acceleration is therefore constant. This means that the force (F = ma) is constant and hence*independent*of time. [Option (e)].**(3)**A body of mass 5 kg at rest is acted on by two forces 3 N and 4 N which are at right angles to each other. The kinetic energy of the body at the end of 10 s is

(a) 250 J

(b) 350 J

(c) 500 J

(d) 750 J

(e) 100 J

The resultant force on the body is √(3

^{2}+ 4^{2}) = 5 N.The acceleration of the body,

*a = F/m =*5 N/ 5 kg = 1 ms^{–2}The velocity of the body at the end of 10 s is given by

*v = v*with usual notations so that

_{0}+ at*v*= 0 + 1×10 = 10 ms

^{–1}

Therefore, the kinetic energy of the body,

*K =*½*mv*^{2}= ½ ×5 ×10^{2}= 250 J.**(4)**A steel ball bearing thrown vertically down bounces (from the floor) 5 m higher than its original level. With what speed was the ball bearing thrown down? Assume elastic collision at the floor and g = 10 ms

^{–.2}

(a) 5 ms

^{–1}(b) 10 ms

^{–1}(c) 12 ms

^{–1}(d) 16 ms

^{–1}(e) 20 ms

^{–1}Since the collision at the floor is elastic, the ball bearing will return to the original level with the magnitude of its velocity unchanged, but the direction reversed. Therefore, the problem is equivalent to finding the velocity (

*v*) with which the ball bearing is to be thrown vertically upwards to reach a height of 5 m. Thus

*v =*√(2*gh*) = √(2×10×5) = 10 ms^{–1}[You can get the above relation by equating the initial kinetic energy to the gravitational potential energy at the maximum height

*h*as ½*mv*^{2 }=*mgh*or from the equation of motion, 0 =*v*^{2}– 2*gh*].**(5)**When a long spiral spring is stretched by 2 cm, its potential energy is

*U*. If this spring is stretched by 3 cm, its potential energy will be

(a) 3

*U/*2(b) 9

*U/*4(c) 5

*U/*2(d) 3

*U*(e) 3

*U/*2^{}We have

*U =*½*K*×(0.02)^{2}= 0.0002*K*where*K*is the spring constant.When the stretch is 3 cm, the potential energy is ½

*K*×(0.03)^{2}= 0.00045*K*= 9*U*/4[It will be better to use the proportionality relation in situations like this:

*U*α 2

^{2}

*X*α 3

^{2}

Therefore,

*U/X =*4/9 so that*X =*9*U/*4]**(6)**The momentum of a car is increased by 3%. Because of this the kinetic energy of the car will be

(a) increased by 3%

(b) decreased by 3%

(c) increased by 6%

(d) decreased by 6%

(e) increased by 9%

The kinetic energy is directly proportional to the square of the momentum (

*K = p*^{2}/2*m*), Therefore, when the*percentage**increase*in the kinetic energy will be*twice*the*percentage**increase*in the momentum [Option (c)].[The fractional change in kinetic energy is given by D

*K/K =*2D*p/p*which is twice the fractional change in momentum. The*percentage change*in kinetic energy is therefore twice the*percentage**change*in the momentum].

*The following questions are for AP Physics C aspirants:***(7)**A body suffers a displacement

*r*under the action of a force

*F*

**which varies inversely**

**as the displacement**

*r*. The work done by the force is proportional to

(a)

*r*^{0}(b)

*r*(c)

*r*^{2}(d) 1/

*r*(e) ln

*r*We have

*W =*∫**F.dr**Since the force

**F**and the displacement**r**are in the same direction*W =*∫

*Fdr*

But

*F***=***k*/*r*where*k*is a constant.Therefore,

*W =*∫*Fdr =*∫*kdr/r*=*k*ln*r*Thus the work done is proportional to the logarithm of displacement [Option (e)].

**(8)**A conservative force

**F**= – 4(x

**î**+ y

**ĵ**) newton moves a particle in the xy-plane. Initially the particle is at the origin. The work done by the force when the particle is given a displacement

**r**= (2

**î**+ 2

**ĵ)**metre is (

**î**and

**ĵ**are unit vectors along the x and y directions respectively)

(a) – 4 J

(b) – 8 J

(c) 8 J

(d) –16 J

(e) 16 J

You may imagine the particle to be moved first along the positive x-direction from the origin O to the point (2, 0) and then along the positive y-direction from the the point (2, 0) to the point (2,2) as shown in the figure. The value of x changes from zero to 2 during the motion along the x-direction and the value of y changes from zero to 2 during the motion along the y-direction.

Work done during the motion along the x-direction

*is**W*

_{1}

*=*

_{0}∫

^{2}

*F*

_{x}

*dx*=

_{0}∫

^{2}(– 4x)

*dx*= (– 4)×(2

^{2}/2) = – 8 J.

Work done during the motion along the y-direction

*is**W*

_{2}

*=*

_{0}∫

^{2}

*F*

_{y}

*dy*=

_{0}∫

^{2}(– 4y)

*dy*= (– 4)×(2

^{2}/2) = – 8 J.

Therefore, total work done = – 16 J.

In the next post we will discuss more questions in this section.

## No comments:

## Post a Comment