Impulse-Momentum of Particles
Linear and Angular Momentum
Learning Objectives:
- Relate mass, velocity, and linear momentum for a moving object, and calculate the total linear momentum of a system of objects.
Linear Momentum:
- Linear Momentum of a moving object is defined as the product of the mass and velocity. \[\mathrm{Linear\ Momentum} = m \times v\]
- Linear Momentum is a vector quantity.
- It has both a magnitude (mass \(\times\) speed) and a direction (the direction the object is moving).
Angular Momentum:
- Angular Momentum of a moving object is defined as the product of the Moment of Inertia and angular velocity. \[\mathrm{Angular\ Momentum} = I \times \omega\]
- Angular Momentum is a vector quantity.
- It has both a magnitude (moment of Inertia \(\times\) angular velocity) and a direction according to the right hand rule.
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
Solved Example: 32-1-02
A 10 kg cart is moving to the right with a velocity of $5 \, \text{m/s}$. A 2 kg ball is thrown to the left with a velocity of $15 \, \text{m/s}$ relative to the ground. What is the final velocity of the cart if the ball is caught by the cart?
A. $1.67 \, \text{m/s}$
B. $3.15 \, \text{m/s}$
C. $4.08 \, \text{m/s}$
D. $5.61 \, \text{m/s}$
\[\text{Initial momentum of the cart} = m_{\text{cart}} \cdot v_{\text{cart}} = 10 \, \text{kg} \cdot 5 \, \text{m/s} = 50 \, \text{kg} \cdot \text{m/s}\] \[\text{Initial momentum of the ball} = m_{\text{ball}} \cdot v_{\text{ball}} = 2 \, \text{kg} \cdot (-15 \, \text{m/s}) = -30 \, \text{kg} \cdot \text{m/s}\] \[\text{Total initial momentum} = 50 \, \text{kg} \cdot \text{m/s} - 30 \, \text{kg} \cdot \text{m/s} = 20 \, \text{kg} \cdot \text{m/s}\] \[ \text{Total mass after catching the ball} = m_{\text{cart}} + m_{\text{ball}} = 10 \, \text{kg} + 2 \, \text{kg} = 12 \, \text{kg}\] \[\text{Total initial momentum} = \text{Total final momentum}\] \[20 \, \text{kg} \cdot \text{m/s} = 12 \, \text{kg} \cdot v_{\text{final}}\] \[v_{\text{final}} = \dfrac{20 \, \text{kg} \cdot \text{m/s}}{12 \, \text{kg}} = \dfrac{20}{12} \approx 1.67 \, \text{m/s}\]
Correct Answer: A
Solved Example: 32-1-03
A gun of mass 3000 kg fires horizontally a shell of mass 50 kg with a velocity of 300 m/s. What is the velocity with which the gun will recoil?
A. -5 m/s
B. 10 m/s
C. 50 m/s
D. 30 m/s
Intial momentum = Final momentum \begin{align*} 0 &= m_g v_g + m_s v_s\\ m_g v_g &= - m_s v_s \end{align*} \begin{align*} v_g &= \dfrac{- m_s v_s}{m_g}\\ &= \dfrac{- 50 \times 300}{3000}\\ &= - 5\ \mathrm{m/s} \end{align*}
Correct Answer: A
Impulse-Momentum Equation
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. \begin{align*} F &= m.a\\ F &= m. \frac{\Delta v}{\Delta t}\\ F. \Delta t &= m. \Delta v \end{align*} Or, when instantaneous quantities are considered,
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, \begin{align*} F\Delta t &= 20 \times 0.05\\ &= 1\ \mathrm{N.s} \end{align*}
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, \begin{align*} m \Delta v &= 0.2 [25-(-25)]\\ &= 10\ \mathrm{N.s} \end{align*}
Correct Answer: C
Solved Example: 32-2-05
A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s$^2$) nearly:
A. 0 kg.m/s
B. 4.2 kg.m/s
C. 2.1 kg.m/s
D. 1.4 kg.m/s
\begin{align*} \mathrm{Impulse} &= \mathrm{Change\ in\ momentum}\\ &= - mv_1 - mv_1\\ &= -m ( v_1 + v_1)\\ &= 0.15 (14.14 + 14.14)\\ &= 4.2\ \mathrm{kg.m/s} \end{align*}
Correct Answer: B
Solved Example: 32-2-06
Which of the following statements about momentum is true?
A. Momentum is a scalar quantity.
B. Momentum is conserved in all collisions.
C. Momentum cannot change in a closed system.
D. None of the above.
Correct Answer: B
Solved Example: 32-2-07
Which of the following represents impulse?
A. Force multiplied by distance
B. Change in momentum
C. Mass times acceleration
D. All of the above
Correct Answer: B
Solved Example: 32-2-08
The area under a force vs. time graph represents:
A. Work done
B. Kinetic energy
C. Impulse
D. Momentum
Correct Answer: C
Solved Example: 32-2-09
Which of the following does NOT affect the impulse experienced by an object?
A. The magnitude of the force applied.
B. The duration of the force applied.
C. The mass of the object.
D. The angle at which the force is applied.
Correct Answer: C
Solved Example: 32-2-10
What is the unit of impulse?
A. Joules
B. Newtons
C. Newton-seconds
D. Kilograms
Correct Answer: C
Impact
Learning Objectives:
- To apply conservation of momentum in simple situations.
- Describe the concept of coefficient of restitution.
- To understand the basic ideas of elastic and inelastic collisions.
- Apply the definition of the coefficient of restitution to compute the velocity of an object after impact.
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, \[e = \dfrac{\mathrm{Relative\ velocity\ of \ separation}}{\mathrm{Relative\ velocity\ of\ approach}}\]
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 as well as momentum is conserved.
If e = 0, perfectly plastic collision. In such case energy is not conserved, but momentum is conserved.
No-w-ay in collaboration with H. Caps, CC BY-SA 4.0, via Wikimedia Commons
Solved Example: 28-3-05
A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits ground and bounces off ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is:
A. 0.46
B. 0.55
C. 0.59
D. 0.64
Velocity just before impact with the ground,
\begin{align*}
v^2 &= u^2 + 2gs\\
v^2 &= 0 + 2\times 9.81 \times 1\\
v &= 4.43\ \mathrm{m/s}
\end{align*}
It is given that the velocity after impact is 80%
\[v_{\mathrm{after\ impact}} = 0.8 \times 4.43 = 3.54\ \mathrm{m/s}\]
This is now taken as u for return (upward) motion. The final velocity, at the topmost point, will be zero.
\begin{align*}
v^2 &= u^2 + 2gs\\
0 &= 3.54^2 + 2 (-9.81) s\\
s &=0.64\ \mathrm{m}
\end{align*}
Correct Answer: D
Solved Example: 32-3-01
A ball is dropped from a height 'y' above a smooth floor. How high will it 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 impacts at an angle of 32.5$^\circ$ from the horizontal on a smooth floor. At what angle will it rebound if the coefficient of between the ball and the floor is 0.30?
A. 11.3$^\circ$
B. 8.6$^\circ$
C. 9.1$^\circ$
D. 10.8$^\circ$
Resolve the velocity of the ball into horizontal and vertical components.
The horizontal velocity will NOT be affected by impact, as there is no friction. (smooth floor)
$v \cos \theta = v \cos 32.5^\circ$ will remain same.
\[v'_{\mathrm{horizontal}} = v \cos 32.5^\circ = 0.84 v\]
The vertical velocity will change due to impact.
\begin{align*} e &=- \left[\dfrac{v'_{\mathrm{vertical}} - 0}{-v\sin \theta - 0}\right]\\ 0.30 &= \dfrac{v'_{\mathrm{vertical}}}{v\sin 32.5^\circ}\\ v'_{\mathrm{vertical}} &= 0.30 \times v\sin 32.5^\circ\\ v'_{\mathrm{vertical}} &= 0.16 v \end{align*} Final velocity after impact will be resultant of these two velocities and the new angle can be found out as: \begin{align*} \tan \theta &= \dfrac{v'_{\mathrm{vertical}}}{v'_{\mathrm{horizontal}}}\\ &= \dfrac{0.16 v}{0.84 v}\\ &= 0.19 \end{align*} \[\theta = \tan^{-1} (0.19) = 10.74^\circ\]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
Solved Example: 32-3-06
A rigid ball of mass m strikes a rigid wall at 60$^\circ$ and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be:
A. mV
B. 2mV
C. $\dfrac{mV}{2}$
D. $\dfrac{mV}{3}$
The vertical component will not be affected, so there is no impulse in y-direction.
The horizontal component will be reversed, so there is impulse in x-direction, which is given by:
\begin{align*}
\mathrm{Impulse} &= \mathrm{Net\ change\ in\ momentum}\\
&= mV \cos \theta - (- mV \cos \theta)\\
&= 2mV \cos \theta\\
&= 2mV \cos 60^\circ\\
&= mV
\end{align*}
Correct Answer: A
Solved Example: 32-3-07
Which of these statements is correct for inelastic collision?
A. The kinetic energy of a body during impact remains constant.
B. The kinetic energy of a body before impact is equal to the kinetic energy of a body after impact.
C. The kinetic energy of body before impact is less than the kinetic energy of a body after impact.
D. The kinetic energy of body before impact is more than the kinetic energy of a body after impact.
In collision, the linear momentum is always conserved.
However, the energy is conserved only when the collision is perfectly elastic. For inelastic collisions, some energy is lost due to molecular bonding takes place. This means for inelastic collision, the kinetic energy of body before impact is more than the kinetic energy of a body after impact.
Correct Answer: D
Solved Example: 32-3-08
In a completely inelastic collision:
A. Momentum is conserved.
B. Kinetic energy is conserved.
C. Objects stick together after the collision.
D. Both A and C.
Correct Answer: D
Solved Example: 32-3-09
In a perfectly elastic collision:
A. Momentum is conserved.
B. Kinetic energy is conserved.
C. Both A and B are true.
D. None of the above.
Correct Answer: C
Solved Example: 32-3-10
During a collision, which of the following remains constant?
A. Kinetic energy
B. The total energy of the system
C. The velocity of the objects
D. The impulse experienced by each object
Correct Answer: B