A ball of mass 10 kg is moving with a velocity of 10 m/s. It strikes another ball of mass 4 kg, 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
A
12 m/s,10 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12 m/s,25 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6 m/s,12 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2 m/s,20 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C6 m/s,12 m/s Let the their velocities after the collision be v1 and v2. As we know for elastic collision.
Relative velocity of approach = relative velocity of speration 10−4=v2−v1⇒6=v2−v1 ⇒v1=v2=6
Applying conservation of momentum, 10×10+5×4=10v1+5v2 120=10v1+5v2 120=10(v2−6)+5v2=15v2−60 15v2=180⇒v2=12 cm/sec, v1=6 cm/sec