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A bullet of mass 50 gm is fired with a velocity of 90 m/s towards a ball of mass 500 gm, initially at rest. When the bullet strikes the ball, it gets embedded into it and the system starts moving. What is the ratio of the kinetic energy of the bullet to the kinetic energy of the (bullet + ball) system?


Your Answer
A
10.5
Correct Answer
B
11
Your Answer
C
8
Your Answer
D
14

Solution

The correct option is D 11
Initially, K.Ebullet=12×m×v2
=12×(50×103)×(90)2=202.5 J
Let velocity of system after collision be v.
msystem=0.05+0.5=0.55 kg

Applying the principle of conversation of momentum to the system,
Pi=Pf
(0.05×90)+0=(0.05+0.5)v
v=4.50.55=8.18 m/s

and Kinetic energy of the system
K.Esystem=12×(0.5+0.05)×(8.18)2=18.4 J
Ratio of the kinetic energies:
K.EbulletK.Esystem=202.518.4=11

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