Work, Power and EnergyHard
Question
A ball of mass 10 kg is moving with a velocity of 10 m/ s. it strikes another ball of mass 5kg, which is moving in the same direction with a velocity of 4 m/s. If the collision is elastic their velocities after collision will be respectively :
Options
A.12m/s
B.12m/s,25m/s
C.6m/s,12m/s
D.8m/s,20m/s
Solution
Let their velocities after the collision be v1 and v2. As we know for elastic collision.
Relative velocity of approach
= relative velocity of separation
10 - 4 = v1 - v2 ⇒ 6 = v2 - v1
⇒ v1 = v2 - 6
Applying conservation of momentum,
10 × 10 + 5 × 4 = 10 v1 + 5v2
120 = 10 v1 + 5v2
120 = 10(v1 - 6) + 5v2 = 15v2 - 60
15v2 = 180 ⇒ v2 = 12cm/sec.
v1 = 6cm/sec
Relative velocity of approach
= relative velocity of separation
10 - 4 = v1 - v2 ⇒ 6 = v2 - v1
⇒ v1 = v2 - 6
Applying conservation of momentum,
10 × 10 + 5 × 4 = 10 v1 + 5v2
120 = 10 v1 + 5v2
120 = 10(v1 - 6) + 5v2 = 15v2 - 60
15v2 = 180 ⇒ v2 = 12cm/sec.
v1 = 6cm/sec
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