A bullet of mass m= 0.05kg is fired along an incline with a speed of 350m/s. It
ID: 1450738 • Letter: A
Question
A bullet of mass m= 0.05kg is fired along an incline with a speed of 350m/s. It strikes a block of wood of mass M= 1.50kg placed on the incline as shown above. In case#1 the bullet passes through the block completely and emerges with a speed half its initial value. In case#2 the bullet is imbedded into the wooden block. a) For case#1, find the speed of the wooden block immediately after the bullet has passed through it. b) For case#1, find the height H to which the wooden block would rise after the impact? c)or case #2, what is the speed of the wooden block (with the bullet inside it) soon after impact? d) For case #2, what is the impulse imparted to the wooden block by the bullet during the impact? e) In case#2 if the duration of the impact is 0.05s, what is the average force exerted on the bullet by the block during impact?Explanation / Answer
a) Apply conservation of momentum
m*u = m*u/2 + M*V
==> V = (m*u - m*u/2)/M
= (0.05*350 - 0.05*350/2)/1.5
= 5.83 m/s
b) Apply conservation of energy
M*g*H = (1/2)*M*V^2
H = V^2/(2*g)
= 5.83^2/(2*9.8)
= 1.736 m
c) again Apply conservation of momentum
m*u = (m + M)V
==> V = m*u/(m+M)
= 0.05*350/( 0.05 + 1.5)
= 11.3 m/s
d) Impulse on wooden block = change in its momentum
= M*(V-0)
= 1.5*11.3
= 16.9 N.s or kg.m/s
e) Apply, Impulse = Favg*dt
==> Favg = Impulse/dt
= 16.9/0.05
= 338.7 N
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