Question #3847

As everyone knows, bullets bounce from Superman’s chest. Suppose Superman, mass
104 kg, while not moving, is struck by a 4.2 g bullet moving with a speed of 835 m/s. If the
collision is elastic, find the speed that Superman had after the collision. (Assume the
bottoms of his superfeet are frictionless.)

Expert's answer

Let m_{1} be the mass of the bullet (m_{1} = 4.2 g), m_{2} - the mass of Superman (m_{2} = 104 kg), .U - speed of the bullet before the colision,

V - speed of Superman after the collision.

Consider that the bullet stopped after the collision.

According to the law of conservation of linear momentum:

m_{1}U = m_{2}VV = U*m_{1}/m_{2}V = 835*0.0042/104 = 0.0337 m/s

So, the speed of Superman is 0.0337 m/s.

V - speed of Superman after the collision.

Consider that the bullet stopped after the collision.

According to the law of conservation of linear momentum:

m

So, the speed of Superman is 0.0337 m/s.

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