Learning Objectives:

• Relate mass, velocity, and linear momentum for a moving object, and calculate the total linear momentum of a system of objects.

Formally, the momentum of a moving object is defined as the product of the mass and velocity. Note that it is not just the speed, rather the velocity. Remember, the difference between the speed and the velocity is that velocity is a vector, it has a direction associated with it. This means that momentum is a vector quantity. It has both a magnitude (mass \(\times\) speed) and a direction (the direction the object is moving).

Solved Example: 32-1-01

Momentum possessed by spinning objects is called:

A. Linear momentum

B. Angular momentum

C. Normal momentum

D. Degrees' momentum

The keyword here is 'spinning' object, so it has a rotational motion. Hence it must possess angular momentum.

Correct Answer: B

Learning Objectives:

• To understand and use the impulse-momentum theorem

The impulse applied by the net force on a system is equal to the change of momentum of the system. \[F = m.a\] \[F = m. \frac{\Delta v}{\Delta t}\]

\[F. \Delta t = m. \Delta v\]

Therefore, Impulse = change in momentum
If force is variable and if we know graph of Force versus time, then impulse is area under the F-t diagram.
If there is no force applied on the system then momentum of the system is conserved in magnitude and direction.

Solved Example: 32-2-01

Impulse is equal to:

A. Force $\times$ time

B. Change in momentum

C. A or B

D. None of the above

Correct Answer: C

Solved Example: 32-2-02

_______ is defined as the integral of force with respect to time.

A. Momentum

B. Impulse

C. Velocity

D. Acceleration

Correct Answer: B

Solved Example: 32-2-03

If a 10-kg object experiences a 20-N force for a duration of 0.05-second, then what is the momentum change of the object?

A. 1 N.s

B. 400 N.s

C. 0.5 N.s

D. 200 N.s

Impulse can be calculated by, \[F\Delta t = 20 \times 0.05 = 1\ N.s\]

Correct Answer: A

Solved Example: 32-2-04

When hit, the velocity of a 0.2 kg baseball changes from +25 m/s to -25 m/s. What is the magnitude of the impulse delivered by the bat to the ball?

A. 1 N.s

B. 5 N.s

C. 10 N.s

D. 20 N.s

Impulse can be calulated by, \[m \Delta v = 0.2 (25-(-25) = 10\ N.s\]

Correct Answer: C

Learning Objectives:

• To apply conservation of momentum in simple situations.

• Describe the concept of coefficient of restitution

• Apply the definition of the coefficient of restitution to compute the initial upward velocity of an object after impact.

• To understand the basic ideas of elastic and inelastic collisions.

• Apply linear momentum conservation to two-dimensional collisions.

\[m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'\] where,
\(m_1 , m_2\) = masses of two bodies
\(v_1, v_2\) = the velocities of two bodies just before impact
\(v_1', v_2'\) = the velocities of two bodies just after impact
The coefficient of restitution is defined as, \[e= \dfrac{\mathrm{Relative\ velocity\ of \ separation}}{\mathrm{Relative\ velocity\ of\ approach}}\]

\[e= \dfrac{(v_2')_n - (v_1')_n}{(v_1)_n - (v_2)_n}\]

momentum of the system is conserved in both elastic and inelastic collisions however; kinetic energy is conserved only in the elastic collisions.
If e = 1, perfectly elastic collision. In such case energy is also conserved and momentum is conserved.
If e = 0, perfectly plastic collision. In such case energy is not conserved, but momentum is conserved.

Solved Example: 32-3-01

A ball is dropped from a height y above a smooth floor. How high will rebound if the coefficient of restitution between the ball and the floor is 0.60?

A. 0.45y

B. 0.40y

C. 0.60y

D. 0.36y

When a ball is dropped from a height y, its velocity at the time of impact will be $\sqrt{2gy}$.

Since coefficient of restitution is not 1, some energy is lost and final velocity will not be same.

\[\mathrm{Coefficient\ of\ restitution} = \dfrac{\mathrm{Relative\ velocity\ of\ separation}}{\mathrm{Relative\ velocity\ of\ approach}}\] Ground remains stationary all the time.

\[0.6 = \dfrac{v'-0}{\sqrt{2gy}-0}\] \[v' = 0.6 \sqrt{2gy}\]
This becomes 'u' for the return journey.
Use newton's equation of motion. \begin{align*} v^2 &= u^2 + 2as\\ 0 &= 0.36 (2gy) + 2gs\\ s &= 0.36y\\ \end{align*}

Correct Answer: D

Solved Example: 32-3-02

A ball is thrown at an angle of 32.5$^\circ$ from the horizontal towards a smooth floor. At what angle will it rebound if the coefficient of between the ball and the floor is 0.30?

A. 11.33$^\circ$

B. 8.67$^\circ$

C. 9.12$^\circ$

D. 10.82$^\circ$

$v^{2}=u^{2}+2as, v=4.43\sqrt {y}, e=\dfrac {v_{2}}{v_{1}}, v_{2}=2.656\sqrt {y}$

This becomes 'u' for return journey.

$v^{2}=u^{2}+2as, \mathrm{\ or,\ } s=0.36m$

Correct Answer: D

Solved Example: 32-3-03

A tennis ball is dropped into a cement floor from a height of 2 m. It rebounds to a height of 1.8 m. What fraction of energy did it lose in the process of striking the floor?

A. One-tenth

B. One-fourth

C. One-third

D. One-seventh

Initial PE = mgh = mg(2)

Final PE = mgh = mg(1.8)

Energy lost = mg(0.2)

Fraction of Energy lost = $\dfrac{mg(0.2)}{mg(2)}$ = 0.1

Correct Answer: A

Solved Example: 32-3-04

The coefficient of restitution of a perfectly plastic impact is:

A. 0

B. 1

C. 2

D. 3

From the Newton's Law of collision of Elastic bodies. Velocity of separation = e $\times$ Velocity of approach. \[(v_2- v_1) = e(u_1- u_2)\] Where e is a constant of proportionality and it is called the coefficient of restitution and its value lies between 0 to 1. The coefficient of restitution of a perfectly plastic impact is zero, because all the K.E. will be absorbed during perfectly plastic impact.

Correct Answer: A

Solved Example: 32-3-05

Collisions in which objects rebound with the same speed as they had prior to the collision are known as:

A. Elastic collisions

B. Inelastic collisions

C. Static collisions

D. Plastic collisions

If e = 1, perfectly elastic collision. In such case energy is also conserved and momentum is conserved. In this situation, the body rebounds with the same speed after hitting the ground.
On the other extreme, if e = 0, perfectly plastic collision. In such case energy is not conserved, but momentum is conserved. The body gets stuck and relative velocity of separation is zero.

Correct Answer: A