A 12.2 g bullet is fired into a 59 g wooden block initially at rest on a horizon

Question

A 12.2 g bullet is fired into a 59 g wooden block initially at rest on a horizontal surface. The acceleration of gravity is 9.8 m/s2 . After impact, the block slides 4.18 m before coming to rest.

If the coefficient of friction between block and surface is 0.292 , what was the speed of the bullet immediately before impact? Answer in units of m/s.

A 12.2 g bullet is fired into a 59 g wooden block initially at rest on a horizontal surface. The acceleration of gravity is 9.8 m/s2 . After impact, the block slides 4.18 m before coming to rest.

If the coefficient of friction between block and surface is 0.292 , what was the speed of the bullet immediately before impact? Answer in units of m/s.

Explanation / Answer

mass of the bullet is m = 12.2 g = 0.0122 kg
mass of the wooden block is  M = 59 g = 0.059 kg

distance traveled by the block is d = 4.18 m

If the body is moving in the rough horizontal plane then the acceleration is a = - g

Let the bullet is immersed in the block then the both block and bullet is moving together, then from the equation of motion,

                            V2 - U2 = 2ad

                             0   - U2 = 2(-g) d

                                      U=  2gd   = (2*0.292*9.8*4.18)

                                          = 4.89 m/s

This is the velocity of the combined mass

From the conservation of momentum the

                        mV = (m+M)U

                             V = (m+M)U/m

= (0.0122 + 0.059) * 4.89 / 0.0122

V = 28.538 m/s (Ans)

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