A(n) 9.9 g bullet is fired into a(n) 138 g block of wood at rest on a horizontal

Question

A(n) 9.9 g bullet is fired into a(n) 138 g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 13 m before coming to rest. The acceleration of gravity is 9.8 m/s2 . If the coefficient of friction between the surface and the block is 0.6, find the speed of the bullet before impact. Answer in units of m/s. A(n) 9.9 g bullet is fired into a(n) 138 g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 13 m before coming to rest. The acceleration of gravity is 9.8 m/s2 . If the coefficient of friction between the surface and the block is 0.6, find the speed of the bullet before impact. Answer in units of m/s. A(n) 9.9 g bullet is fired into a(n) 138 g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 13 m before coming to rest. The acceleration of gravity is 9.8 m/s2 . If the coefficient of friction between the surface and the block is 0.6, find the speed of the bullet before impact. Answer in units of m/s.

Explanation / Answer

mass of bullet (m) = 9.9 gram
Speed of the bullet initially = u
Mass of wooden block (M) = 138 gram
Therefore initial momentum = mu +M*0
= mu ---------(1)
Let say after the collision both bullet and block has the speed = V
Final momentum after the collision = (m+M)V ----------(2)
Final kinetic energy of the system = (1/2)*(m+M)V2 -----------(3)
Now this energy will be dissipiated due to friction
= friction force*distance (S)
= [u*(m+M)g]*13 -----------(4)
Now applying energy conservation by equating 3 and 4
(1/2)*(m+M)V2 =  [u*(m+M)g]*13
(1/2)V2 = ug*13
(1/2)V2 = 0.6*9.8*13
V = 12.364 m/s
Now putting this in equation 2 , therefore
final momentum = (m+M)V = (9.9+138)*12.364
Now applying the conservation of momentum
mu = (9.9+138)*12.364
9.9*u = (9.9+138)*12.364
u = 184.71 m/s
Hence the speed of the bullet before the impact is 184.71 m/s

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