1. A 500-gram object, A, moving at 10 m/s and 200-gram object, B, moving at 12 m/s. The objects approach each other and collide. If the speed of object A after the collision is 6 m/s, what is the speed of object B after the collision?

__Known :__

Mass of object 1 (m_{1}) = 500 gram = 0.5 kg

Mass of object 2 (m_{2}) = 200 gram = 0.2 kg

Initial velocity of object 1 (v_{1}) = -10 m/s

Initial velocity of object 2 (v_{2}) = 12 m/s

The final velocity of object 1 (v_{1}’) = 6 m/s

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

__Wanted__ : the final velocity of object 2 (v_{2}’)

__Solution :__

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

(0.5)(-10) + (0.2)(12) = (0.5)(6) + (0.2)(v_{2}’)

-5 + 2.4 = 3 + 0.2 v_{2}’

-2.6 = 3 + 0.2 v_{2}’

-2.6 – 3 = 0.2 v_{2}’

-5.6 = 0.2 v_{2}’

v_{2}’ = -5.6 / 0.2

v_{2}’ = -28 m/s

The speed of object 2 after collision is 28 m/s.

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

[irp]

2. Two equal-mass objects approach each other and collide. The speed of object 1 is 6 m/s and the speed of object 2 is 8 m/s. After the collision, object 2 moves leftward with speed of 5 m/s. What is the magnitude and direction of the velocity of object 1 after the collision?

__Know :__

Mass of object 1 (m_{1}) = m

Mass of object 2 (m_{2}) = m

Initial speed of object 1 (v_{1}) = -6 m/s

initial speed of object 2 (v_{2}) = 8 m/s

The final speed of object 2 (v_{2}’) = -5 m/s

__Wanted__ : the magnitude and direction of object 1 after collision

__Solution :__

Formula of conservation of linear momentum :

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

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

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

v_{1} + v_{2} = v_{1}’ + v_{2}’

-6 + 8 = v_{1}’ – 5

2 = v_{1}’ – 5

2 + 5 = v_{1}’

v_{1}’ = 7 m/s

The speed of object 1 after collision (v_{1}’) is 3 m/s.

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

[irp]

3. A 1-kg ball 1 and 2-kg ball 2 have the same direction and collide inelastically. Before the collision, ball 1 moves with speed of 10 m/s and ball 2 moves with speed of 5 m/s. The speed of ball 2 after the collision is 4 m/s. Determine the speed of ball 1 after the collision.

__Known :__

Mass ball 1 (m_{1}) = 1 kg

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

The speed of ball 1 (v_{1}) = 10 m/s

The speed of ball 2 (v_{2}) = 5 m/s

The final speed of ball 2 (v_{2}’) = 4 m/s

*The plus sign of the velocity indicates that the balls have the same direction.*

__Wanted ____:__ the final velocity of ball 1 (v_{1}’)

__Solution :__

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

(1)(10) + (2)(5) = (1)(v_{1}’) + (2)(4)

10 + 10 = v_{1}’ + 8

20 – 8 = v_{1}’

v_{1}’ = 12 m/s

The speed of ball 1 after collision is 12 m/s.

[wpdm_package id=’1190′]

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