A 50 gram Bullet moving with a velocity 10 metre per second strikes a block of mass 950 gram at rest and gets embedded in it the loss in kinetic energy will be
a. 100% b. 95% c. 5% d. 50%
Answers
The loss of Kinetic Energy will be, 95 %
Explanation:
Given data,
The mass of the bullet, m = 50 g
= 0.05 k
The mass of the block, M = 950 g
= 0.95 kg
The velocity of the bullet, u = 10 m/s
The velocity of the block, U = 0
According to the law of conservation of momentum
mu = (M + m) v
v = mu /(M + m)
= 0.05 x 10 / ( 0.95 + 0.05)
= 0.5 m/s
The K.E of the bullet,
K.E = ½ mu²
= ½ x 0.05 x 10²
= 2.5 J
The kinetic energy of the combination,
K.E = ½ (M + m) v²
= ½ (1) 0.5²
= 0.125 J
The percentage of K.E of the combination to the bullet is,
K. E % = 0.125 J x (100 / 2.5) %
= 5 %
Therefore, the loss of K.E will be, 95 %