 # Inelastic collisions in one dimension – problems and solutions

1. A 30-gram bullet moving at 30 m/s collide a 1-kg block at rest. Determine the speed of the block if the bullet and the block lock together as a result of the collision.

Known :

Mass of bullet (m1) = 30 gram = 0.03 kg

Mass of block (m2) = 1 kg

Initial speed of bullet (v1) = 30 m/s

initial speed of block (v2) = 0 (block at rest)

Wanted : the speed of bullet and block after collision (v’)

Solution :

m1 v1 + m2 v2 = (m1 + m2) v’

(0.03)(30) + (1)(0) = (0.03 + 1) v’

0.9 + 0 = 1.03 v’

0.9 = 1,03 v’

v’ = 0.9 / 1,03

v’ = 0.87 m/s

2. Billiard ball A of mass 2 kg moving with speed vA = 6 m.s−1 collides head-on with ball B of equal mass. If the collision is perfectly inelastic, what are the speeds of the two balls after the collision.

Known :

Mass A (mA) = 2 kg

Mass B (mB) = 2 kg

Initial speed of ball A (vA) = -6 m/s

Initial speed of ball B (vB) = 4 m/s

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

Wanted: the speeds of the two balls after the collision. (v’)

Solution :

mA vA + mB vB = (mA’ + mB) v’

(2)(-6) + (2)(4) = (2 + 2) v’

-12 + 8 = 4 v’

-4 = 4 v’

v’ = -4 / 4

v’ = -1 m/s

3. m1 = 3 kg and m2 = 4 kg approach each other in opposite direction with speed of v1 = 10 m/s and v2 = 12 m/s. If the collision is perfectly inelastic, what are the speeds of the two balls after the collision.

Known :

Mass object 1 (m1) = 3 kg

Mass object 2 (m2) = 4 kg

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

The speed of object 2 (v2) = 12 m/s

Wanted : the speed after collision (v’)

Solution :

m1 v1 + m2 v2 = (m1’ + m2) v’

(3)(-10) + (4)(12) = (3 + 4) v’

-30 + 48 = 7 v’

18 = 7 v’

v’ = 18 / 7

v’ = 2.6 m/s

[wpdm_package id=’1157′]