walkthough please! A bullet of mass m=.060 kg hits a 5.000 kg block with a initi
ID: 1371428 • Letter: W
Question
walkthough please! A bullet of mass m=.060 kg hits a 5.000 kg block with a initial speed of 225 m/s. The block is connected to a spring as shown in the Figure below. The friction between the block and the table is negligible. Upon impact the bullet bounces back from the box with a speed of 75 m/s.
a) Calculate the speed of the block right after the collision.
b) As a result of the collision the spring compresses to a maximum of 0.20 m. Find the spring constant.(k)
c) Find the inelastic energy loss during the collision.
Explanation / Answer
a) m1 = 0.06 kg
m2 = 5.00 kg
u1 = 225m/s
u2 =0
v1 = -75m/s
a) m1u1 +m2u2 = m1v1 +m2v2
v2 = (m1u1+m2u2- m2v1)/m2
plugging in the values we get
v2 =3.6 m/s
b) 1/2(m2v22) = 1/2(kx2)
k = m2v22/x2
k = 1620 N/m
c) loss in energy = 1/2 m1u12 - (1/2 m2v22+1/2 m1v12)
= 1317.6 J
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