# Partially elastic collisions

In partially elastic collisions, the law of conservation of momentum is applicable, while the conservation of kinetic energy law is not applicable. At the time a collision takes place, some kinetic energy is converted to sound energy, heat energy, and internal energy. The use of the word elastic signifies that after the collision, the two objects do not stick together but bounce off.

An example of partially elastic collision is the one-dimensional collision of two marbles or two pool balls.

Example question 1.

Objects A and B with masses of 1 kg and 2 kg, respectively move in opposite directions at speeds of 4 m/s and 2 m/s, respectively and collide in a partially elastic collision. If after collision object A moves at a speed of 2 m/s, what is the speed of object B?

Known :

m_{A} = 1 kg, m_{B} = 2 kg, v_{A} = 4 m/s (suppose to the right), v_{B} = -2 m/s (suppose to the left), v_{A }’ = – 2 m/s

Wanted : v_{B} ’

Solution :

m_{A} v_{A} + m_{B }v_{B }= m_{A} v_{A} ’ + m_{B} v_{B} ’

(1 kg)(4 m/s) + (2 kg)(-2 m/s) = (1 kg)(-2 m/s) + (2 kg)(v_{B}’)

4 kg m/s – 4 kg m/s = -2 kg m/s + (2 kg) v_{B}’

0 = – 2 kg m/s + (2 kg)(v_{B}’)

-2 kg m/s = – 2 kg v_{B }’

v_{B}’ = 1 m/s

After collision, object B moves at a speed of 1 m/s (suppose to the right) and object A moves at a speed of 2 m/s (suppose to the left)

Example question 2.

Object A and B with a mass of 4-kg and 5-kg approach each other in the opposite direction, as shown in the figure below. After collision, both objects reversed direction with speed of A = 4 m.s^{-1} and speed of B = 2 m.s^{–}^{1}. What is the speed of object B before the collision?

__Known :__

Mass of object A (m_{A}) = 4 kg

Mass of object B (m_{B}) = 5 kg

Velocity of object A before collision (v_{A}) = 6 m/s

Velocity of object A after collision (v_{A}’) = 4 m/s

Velocity of object B after collision (v_{B}’) = -2 m/s

Plus and minus sign indicates that both objects move in opposite direction.

__Wanted :__ Velocity of object B before collision (v_{B})

__Solution :__

m_{A} v_{A} + m_{B} v_{B} = m_{A} v_{A}’ + m_{B} v_{B}’

(4)(6) + (5)(-2) = (4)(4) + (5)(v_{B}’)

24 – 10 = 16 + 5(v_{B}’)

14 – 16 = 5 (v_{B}’)

-2 = 5 (v_{B}’)

v_{B}’ = -2/5

v_{B}’ = -0.4

Minus sign indicates that object direction after collision is different with direction before collision.

Example question 3.

Two objects with the same mass approach each other as shown in the figure below.

If v_{2}‘ is the speed of the object (2) after the collision to rightward with speed of 5 m.s^{–1}, what is the speed of object one v_{1 }‘ (1) after the collision?

__Known :__

Mass of each object = m

Velocity of object 1 before collision (v_{1}) = 8 m/s

Velocity of object 2 before collision (v_{2}) = 10 m/s

Velocity of object 2 after collision (v_{2}‘) = 5 m/s

__Wanted :__ Speed of object 1 after collision (v_{1}‘)

__Solution :__

m_{1} v_{1}+ m_{2 }v_{2} = m_{1 }v_{1}’ + m_{2 }v_{2}’

m (v_{1 }+ v_{2}) = m (v_{1}’ + v_{2}’)

v_{1} + v_{2} = v_{1}’ + v_{2}’

8 + 10 = v_{1}’ + 5

18 = v_{1}’ + 5

v_{1}’ = 18-5

v_{1}’ = 13 m/s