Partially inelastic collisions in one dimension – problems and solutions

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 (m1) = 500 gram = 0.5 kg

Mass of object 2 (m2) = 200 gram = 0.2 kg

Initial velocity of object 1 (v1) = -10 m/s

Initial velocity of object 2 (v2) = 12 m/s

The final velocity of object 1 (v1’) = 6 m/s

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

Wanted : the final velocity of object 2 (v2’)

Solution :

m1 v1 + m2 v2 = m1 v1’ + m2 v2

(0.5)(-10) + (0.2)(12) = (0.5)(6) + (0.2)(v2’)

-5 + 2.4 = 3 + 0.2 v2

-2.6 = 3 + 0.2 v2

-2.6 – 3 = 0.2 v2

-5.6 = 0.2 v2

v2’ = -5.6 / 0.2

v2’ = -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.

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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 (m1) = m

Mass of object 2 (m2) = m

Initial speed of object 1 (v1) = -6 m/s

initial speed of object 2 (v2) = 8 m/s

The final speed of object 2 (v2’) = -5 m/s

Wanted : the magnitude and direction of object 1 after collision

Solution :

Formula of conservation of linear momentum :

m1 v1 + m2 v2 = m1 v1’ + m2 v2

m v1 + m v2 = m v1’ + m v2

m (v1 + v2) = m (v1’ + v2’)

v1 + v2 = v1’ + v2

-6 + 8 = v1’ – 5

2 = v1’ – 5

2 + 5 = v1

v1’ = 7 m/s

The speed of object 1 after collision (v1’) is 3 m/s.

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

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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 (m1) = 1 kg

Mass ball 2 (m2) = 2 kg

The speed of ball 1 (v1) = 10 m/s

The speed of ball 2 (v2) = 5 m/s

The final speed of ball 2 (v2’) = 4 m/s

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

Wanted : the final velocity of ball 1 (v1’)

Solution :

m1 v1 + m2 v2 = m1 v1’ + m2 v2

(1)(10) + (2)(5) = (1)(v1’) + (2)(4)

10 + 10 = v1’ + 8

20 – 8 = v1

v1’ = 12 m/s

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

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