IEMDaily - Video Lecture Notes

Example

Conservation of energy in elastic collision

Example: A ball A moving with a velocity 5 m/s collides elastically with another identical ball at rest such that the velocity of A makes an angle of

30^{\circ}

with the line joining the centres of the balls. Then what will be the scenario after collision?

Solution:As the collision is elastic so kinetic energy as well as momentum both are conserved.
For momentum conservation,

m v = m v_{A} \cos 30^{0} + m v_{B} \cos θ \Rightarrow v = v_{A} \cos 30^{0} + v_{B} \cos θ . . . . (1)

0 = m v_{A} \sin 30^{0} - m v_{B} \sin θ \Rightarrow 0 = v_{A} \sin 30^{0} - v_{B} \sin θ . . . . . (2)

For KE conservation,

\frac{1}{2} m v^{2} = \frac{1}{2} m (v_{A}^{2} \cos^{2} 30^{0} + v_{B}^{2} \cos^{2} θ)

v^{2} = v_{A}^{2} \cos^{2} 30^{0} + v_{B}^{2} \cos^{2} θ . . . . (3)

and

0 = v_{A}^{2} \sin^{2} 30^{0} + v_{B}^{2} \sin^{2} θ . . . . . (4)

(3) + (4), v^{2} = v_{A}^{2} + v_{B}^{2} . . . . (5)

now squaring (1) and (2) and then adding,

v^{2} = v_{A}^{2} + v_{B}^{2} + v_{A} v_{B} \cos (30^{0} + θ)

using (5),

\cos (30^{0} + θ) = 0 = \cos 90^{0} \Rightarrow θ = 90 - 30 = 60^{o}

Thus A and B will move at right angle after collision.
putting

θ = 60^{0}

in (1) and (2),

v = v_{A} \cos 30^{0} + v_{B} \cos 60^{0} \Rightarrow 5 = \frac{\sqrt{3}}{2} v_{A} + \frac{v_{B}}{2} . . . (6)

and

0 = \frac{v_{A}}{2} - \frac{\sqrt{3}}{2} v_{B} . . . (7)

solving (6) and (7),

v_{A} = \frac{5 \sqrt{3}}{2} m / s

and

v_{B} = \frac{5}{2} m / s

Definition

Velocity of approach and velocity of separation for colliding particles

Example

Elastic collision with infinite mass in two dimensions

Let a body of mass

m_{1}

collide with an infinite mass at rest. Let its velocity be

u_{n}

along the normal before collision and

u_{t}

along the tangent.
Using conservation of momentum in tangential direction,

m_{1} u_{t} = m_{1} v_{1, t}

v_{1, t} = u_{t}

Using conservation of momentum in normal direction,

m_{1} u_{n} = m_{1} v_{1, n} + m_{2} v_{2, n}

For elastic collision, velocity of approach equals the velocity of separation

v_{2, n} - v_{1, n} = u_{n}

Solving,

v_{1, n} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{n}

v_{2, n} = \frac{2 m_{1}}{m_{1} + m_{2}} u_{n}

Using approximation

m_{2} >> m_{1}

v_{1, n} = - u_{1, n}

v_{2, n} = 0

Example

Elastic collision of equal masses in two dimensions

Let a body of mass

m

collide with an object of same mass at rest. Let its velocity be

u_{n}

along the normal before collision and

u_{t}

along the tangent.
Using conservation of momentum in tangential direction,

m u_{t} = m v_{1, t}

v_{1, t} = u_{t}

Using conservation of momentum in normal direction,

m u_{n} = m v_{1, n} + m v_{2, n}

For elastic collision, velocity of approach equals the velocity of separation

v_{2, n} - v_{1, n} = u_{n}

Solving,

v_{1, n} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{n}

v_{2, n} = \frac{2 m_{1}}{m_{1} + m_{2}} u_{n}

Using

m_{2} = m_{1}

v_{1, n} = 0

v_{2, n} = u_{n}

Formula

Conservation of momentum equations of inelastic collisions in two dimnsions

Figure shows a 2-dimensional totally inelastic collision. In this case, the first object, mass

m_{1}

, initially moves along the -axis with speed

v_{i 1}

. On the other hand, the second object, mass

m_{2}

, initially moves at an angle

θ_{i}

to the -axis with speed

v_{i 2}

. After the collision, the two objects stick together and move off at an angle

θ_{f}

to the -axis with speed

v_{f}

. Momentum conservation along the -axis yields

m_{1} v_{i 1} + m_{2} v_{i 2} c o s θ_{i} = (m_{1} + m_{2}) v_{f} c o s θ_{f}

Likewise, momentum conservation along the y-axis gives

m_{2} v_{i 2} s i n θ_{i} = (m_{1} + m_{2}) v_{f} s i n θ_{f}

Example

Problem on Conservation of Momentum

Example: A shell is fired from gun with a velocity of

300 m s^{- 1}

making an angle

60^{0}

with the horizontal. It explodes into two fragments when it reaches the highest position. The ratio of the masses of the two pieces is 1 : 3. If the smaller piece stops immediately after the collision. Find velocity of the other.

Solution:
Velocity at highest point

300 \times c o s 60^{0} = 150

m/s

i.e using momentum conservation

150 \times m = \frac{3 m}{4} \times v

v = 200

m/s

Example

Collision of masses in one dimension

Example: Two identical balls A and B each of mass "m" are moving towards each other at equal speed each of "v". Ball A initially is moving along the positive x direction while ball B is moving along negative x direction. If the collision is perfectly elastic, what will be the impulse received by the ball B?
Solution:Since the collision is perfectly elastic, ball B travels in the backward direction i.e along positive x-axis with the same speed before collision.
The change in velocity becomes

v - (- v) = 2 v

∴

change in momentum

= m (2 v) = 2 m v

Which is the impulse received by the body.

Example

Problems on Multiple Collisions in one dimension

Example: Two balls of same mass each

m

are moving with same velocities

v

on a smooth surface as shown in figure. If all collisions between the mass and with the wall are perfectly elastic then what is possible number of collisions between the bodies and wall together?

Solution:In case of elastic collision the velocity after the collision gets interchanged if masses of colliding bodies are same and the direction of velocity reverses if the one of the mass is very large compared to smaller mass and is at rest.
Using the above concepts the first collision will take place between wall and first ball, the velocity of ball will get reversed. Now the second collision will be between first and second ball where the of velocity will get interchanged now
the first ball will move again towards wall and second will move away from wall.
Now the third collision will take place between wall and ball moving towards it, the velocity of ball will get reversed away from wall. and both balls will move in same direction with constant velocity. Total there will be three collisions.