Question

Ball A has mass 5.0 kg and is moving at -3.2 m/s when it strikes stationary ball B, which has mass 3.9 kg, in a head-on collision. If the collision is completely inelastic, what is the common velocity of balls A and B?

Answer 1

A.) -0.40 m/s

B.) -3.60 m/s

C.) -1.80 m/s

Step-by-step explanation:

By the law of momentum conservation:-

=>m1u1 + m2u2 = m1v1 + m2v2

=>5 x (-3.2) + 0 = 5v1 + 3.9v2

=>1.28v1 + v2 = -4.10 ---------------(i)

for elastic collision:-

=>v1 - v2 = u2 - u1

=>v1 - v2 = 0 - (-3.2)

=>v1 - v2 = 3.2 ------------------(ii)

by (i) + (ii) :-

=>2.28v1 = -0.90

=>v1 = -0.40 m/s

=>v2 = -3.60 m/s

(c) By the law of momentum conservaation:-

=>m1u1+m2u2 = (m1+m2)v

=>5 x (-3.2) + 0 = (5 + 3.9) x v

=>v = -1.80 m/s

Answer 2

Momentum conservation in a completely inelastic collision works like this:
mA*vA+mB*vB=v(mA+mB)

Since B was initially stationary pB=0.
Plugging in numbers gives:
5*(-3.2)=v(5+3.9)
v= -1.79 m/s to the left.