1. A 200-gram ball, A, moving at a speed of 10 m/s strikes a 200-gram ball, B, at rest. What is the speed of ball A and ball B after the collision?

__Known :__

Mass of ball A (m_{A}) = 200 gram = 0.2 kg

Mass of ball B (m_{B}) = 200 gram = 0.2 kg

Speed of ball A before collision (v_{A}) = 10 m/s

Speed of ball B before collision (v_{B}) = 0

Wanted: speed of ball A (v_{A}’) and speed of ball B (v_{B}’) after a collision

__Solution :__

v_{A}’ = v_{B }= 0

v_{B}’ = v_{A }= 10 m/s

2. Two equal-mass objects approach each other, collide and then bounce off. The speed of mass A before the collision is 8 m/s and speed of mass B before the collision is 12 m/s. What is the magnitude and direction of the velocity of mass A and mass B after the collision!

__Known :__

Mass A (m_{A}) = m

Mass B (m_{B}) = m

Speed of mass A before collision (v_{A}) = -8 m/s

Speed of mass B before collision (v_{B}) = 12 m/s

*The plus and minus sign indicates that the objects move in opposite direction.*

Wanted: the magnitude and direction of the velocity of mass A (v_{A}’) and mass B (v_{B}’) after a collision

__Solution :__

In a perfectly elastic collision, if the objects of equal mass then the speed of A after the collision (v_{A}’) = the speed of B before collision (v_{B}) and the speed of B after the collision (v_{B}’) = the speed of A before collision (v_{A}).

If before collision, A moving rightward and B moving leftward, then, after collision, A moving leftward and B moving rightward.

The objects have equal mass and the collision is perfectly elastic so the both objects exchanged speed.

v_{A}’ = -v_{B }= -12 m/s

v_{B}’ = v_{A }= 8 m/s

*The plus and minus sign indicates that the objects move in opposite direction.*

3. A 2-kg object, A, moving at a speed of 10 m/s strikes a 4-kg ball, B, moving at a speed of 20 m/s. If the collision is perfectly elastic, what is the speed of object A and object B after the collision?

__Known :__

Mass A (m_{A}) = 2 kg

Mass B (m_{B}) = 4 kg

Speed of object A before collision (v_{A}) = 10 m/s

Speed of object B before collision (v_{B}) = 20 m/s

__Wanted__ : the speed of object A (v_{A}’) and the speed of object B (v_{B}’) after collision

__Solution __:

In perfectly elastic collision, if the objects have equal mass and approach each other, the speed of the object after collision calculated using this formula :

Speed of object A after collision :

4. A 100-gram moving at 20 m/s strikes a wall perfectly elastic collision. What is the magnitude and direction of object’s velocity after collision.

__Known :__

Mass (m_{A}) = 100 gram = 0.1 kg

Mass of wall (m_{B}) = infinity

The speed of mass before collision (v_{A}) = 20 m/s

The speed of mass after collision (v_{B}) = 0

__Wanted__ : the magnitude and direction of velocity after collision

__Solution :__

If v_{B} = 0, m_{A} very small and m_{2} very big then v_{A}’ = -v_{A} and v_{2}’ = 0.

*The plus and minus sign indicates that the objects move in the opposite direction.*

5. Two balls, A with a mass of 2-kg and ball B with a mass of 5-kg, before and after the collision are shown in the figure below. If the collision is perfectly elastic, what is the speed of ball A before the collision?

__Known :__

Mass of ball A (m_{A}) = 2 kg

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

Speed of ball B before collision (v_{B}) = 2 m/s

Speed of ball A after collision (v_{A}‘) = 5 m/s

Speed of ball B after collision (v_{B}‘) = 4 m/s

*Both balls are move to the right so their speed signed positive.*

__Wanted :__ Speed of ball A before collision (v_{A})

__Solution :__

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

2(v_{A}) + (5)(2) = (2)(5) + (5)(4)

2(v_{A}) + 10 = 10 + 20

2(v_{A}) + 10 = 30

2(v_{A}) = 30 – 10

2(v_{A}) = 20

v_{A }= 20/2

v_{A} = 10 m/s

6. Object A and B travel along horizontal plane, as shown in figure below. If the collision is perfectly elastic and the speed of object B after collision is 15 m/s^{1}, what is the speed of object A after collision.

__Known :__

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

Mass of object B (m_{B}) = 2 kg

Speed of object A before collision (v_{A}) = 12 m/s

Speed of object B before collision (v_{B}) = 10 m/s

Speed of object B after collision (v_{B}‘) = 15 m/s

*Both objects, before and after collision, move to right so their velocity signed positive.*

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

__Solution :__

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

(5)(12) + (2)(10) = (5)v_{A}‘ + (2)(15)

60 + 20 = (5)v_{A}‘ + 30

80 = (5)v_{A}‘ + 30

80 – 30 = (5)v_{A}‘

50 = (5)v_{A}‘

v_{A}‘ = 50/5

v_{A}‘ = 10 m/s

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