A 0.0124-kg bullet is fired horizontally into a 3.63-kg wooden block attached to
ID: 1649995 • Letter: A
Question
A 0.0124-kg bullet is fired horizontally into a 3.63-kg wooden block attached to one end of a massless, horizontal spring (k = 876 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.261 m. What is the speed of the bullet?Explanation / Answer
From the given question,
mass of bullet(m)=0.0124 kg
mass of wooden block(M)=3.63 kg
spring constant(k)=876 N/m
amplitude of oscillation(x)=0.261 m
initial speed of bullet=v
Final speed of bullet-wood system=V
by conservation of momentum,
m(v)+ M(0)=(m+M)V
V=mv/(m+M)
By conservation of energy
(1/2)(m+M)V2=(1/2)(k)x2
(m+M)V2=(k)x2
(m+M)[mv/(m+M)]2=(k)x2
(m2v2)/(m+M)=kx2
v2=(kx2)(m+M)/m2
v2=(876*0..2612)(0.0124+3.63)/(0.0124)2
=1.4136*106
v=1188.95 m/s
Speed of bullet is 1190 m/s (Approx)
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