A bullet of mass m = 0.05kg is fired along an incline with a speed of350m/s. It
ID: 1456086 • Letter: A
Question
A bullet of mass m = 0.05kg is fired along an incline with a speed of350m/s. It strike a block of wood of m= 1 50 kg 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 bullet.s imbedded into the wooden block. For case #1, find the speed of the wooden block immediately after the bullet has passed For case #1, find the height II to which the wooden block would rise after the impact? For case #2, what is the speed of the wooden block (with the bullet inside it) soon after impact? For case #2. what is the impulse imparted to the wooden block by the bullet during the impact? In ease # 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)
Using Momentum Conservation,
Initial Momentum = Final Momentum
0.05 * 350 = 1.5 * v + 0.05 * 350/2
Solving for v
v = 5.83 m/s
(b)
K.E Gained by Block after bullet has passed = 1/2*mv^2
This K.E will get converted to Potential Energy at top of it's height.
so,
m*g*h = 1/2*mv^2
9.8*h = 1/2*5.83^2
h = 1.73 m
Height wooden block would rise, H = 1.73 m
(c)
Using Momentum Conservation,
Initial Momentum = Final Momentum
0.05 * 350 = (1.5 + 0.05) * v
Solving for v
v = 11.3 m/s
(d)
Impulse = Change in Mometum
Impulse = m*vf - m*vi
Impulse = 1.5 * 11.3 Kgm/s
Impulse = 16.95 Kg m/s
(e)
Impulse = F*dt
F = Impulse/dt
F = 16.95/0.05
F = 339 N
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