1. A .006 kg bullet is fired at a speed of 240 m/s into a 1.2 kg pendulum of len
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Question
1. A .006 kg bullet is fired at a speed of 240 m/s into a 1.2 kg pendulum of length .73 m. After the impact, the pendulum swings upward. How high does the pendulum swing, before coming momentarily to rest?
2. A .007 kg bullet is fired at a speed of 340 m/s into a 1.7 block that is resting on a frictionless surface and is attached to a spring with spring constant 300 N/m. After impact, the block with the embedded bullet slides across the table, compressing the spring. How far does the block move, before coming momentarily to rest?
3. A .006 kg bullet is fired at a speed of 300 m/s into a 1.5 kg block that is at rest on a table. After impact, the block with the embedded bullet slides across the table before coming to rest again. The coefficient of kinetic friction between the block and the surface of the table is 0.62. How far does the block slide before coming to rest?
Explanation / Answer
(m+M)*v = M*v
v = MV/(m+M) = (240*0.006)/(1.2+0.006) = 1.194 m/s
PE = KE
(M+m)*g*h = 0.5*(M+m)*V^2
h = V^2/2g = 0.073 m
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2)
(m+M)*v = M*v
v = MV/(m+M) = (340*0.007)/(1.7+0.007) = 1.39 m/s
PE = KE
0.5*K*x^2 = 0.5*(M+m)*V^2
x = 0.1048 m
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3)
(m+M)*v = M*v
v = MV/(m+M) = (300*0.006)/(1.5+0.006) = 1.195 m/s
W = KE
f*x = 0.5*(M+m)*V^2
u*(m+M)*g*x = 0.5*(M+m)*V^2
x = V^2/2ug = 0.117 m
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