# Inelastic Collisions

The conservation of kinetic energy law is not applicable in inelastic collisions. The conservation of momentum law is applicable in inelastic collisions if only no external force acts on the two colliding objects. In an inelastic collision, two objects stick together or are attached to each other after the collision.

Example question 1.

Two objects are of the same mass, namely 1 kg. Object 1 moves on a flat plane at a speed of 10 m/s and collides with object two which is at rest. After the collision, the two objects stick together. What is the speed of the two objects after the collision?

Known :

m1 = 1 kg, m2 = 1 kg, v

1 = 10 m/s, v2 = 0

Wanted : v’

Solution :

m1 v1 + m2 v2 = (m1 + m2) v’

(1 kg)(10 m/s) + 0 = (1 kg + 1 kg) v’

10 kg m/s = (2 kg) v’

v’ = 10 kg m/s : 2 kg = 5 m/s

Read :  Coefficient of performance of the cooling machine

Example question 2. Three blocks move at 3 m.s-1 collide another block at rest.

The collision is inelastic. The order of block’s velocity after the collision, from the largest to the smallest is…

Solution :

Figure 1 :

Final momentum = initial momentum

m1 v1 + m2 v2 = (m1 + m2) v

(4m)(3) + (m)(0) = (4m + m) v

12m + 0 = (5m) v

12m = 5m v

v = 12m / 5m = 12/5 = 2.4 m/s

Figure 2 :

Final momentum = Initial momentum

m1 v1 + m2 v2 = (m1 + m2) v

(m)(3) + (3m)(0) = (m + 3m) v

3m + 0 = (4m) v

3m = 4m v

v = 3m / 4m = 3/4 = 0.75 m/s

Figure 3 :

m1 v1 + m2 v2 = (m1 + m2) v

(m)(3) + (m)(0) = (m + m) v

3m + 0 = (2m) v

3m = 2m v

v = 3m / 2m = 3/2 = 1.5 m/s

Example question 3.

Two balls with mass of m1 = 2 kg and m2 = 1 kg are move in opposite direction with speed of v1 = 2 ms-1 and v2 = 4 ms-1 as shown in figure below. If a collision is inelastic, what is the speed of both balls after the collision?

Known :

Mass of ball 1 (m1) = 2 kg Mass of ball 2 (m2) = 1 kg

Velocity of ball 1 before collision (v1) = 2 m/s

Velocity of ball 2 before collision (v2) = -4 m/s

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

Wanted : Velocity of balls after collision (v’)

Solution :

m1 v1 + m2 v2 = (m1 + m2) v’

(2)(2) + (1)(-4) = (2 + 1) v’

4 – 4 = (3) v’

0 = (3) v’

v’ = 0

Read :  Equation of astronomical telescope

Example question 4.

Two objects, A and B, with a mass of each object, is 1.5-kg approach each other with speed of vA = 4 m.s-1 and vB = 5 m.s-1. If the collision is inelastic, what is the speed of both objects after the collision?

Known :

Mass of object A (mA) = 1.5 kg

Mass of object B (mB) = 1.5 kg

Velocity of object A before collision (vA) = 4 m/s (plus sign, to rightward)

Velocity of object B before collision (vB) = -5 m/s (minus sign, to leftward)

Wanted : The speed of both objects after collision

Solution :

Conservation of linear momentum :

mA vA + mB vB = (mA + mB) v’

(1.5)(4) + (1.5)(-5) = (1.5 +1.5) v’

6 – 7.5 = (3) v’

-1.5 = (3) v’

v’ = -1.5 / 3

v’ = -0.5 m/s