A 9.20 Times 10^-3 kg bullet is fired horizontally into a 2.43-kg wooden block a

Question

A 9.20 Times 10^-3 kg bullet is fired horizontally into a 2.43-kg wooden block attached to one end of a massless, horizontal spring (k = 857 N/m). The other end of the sprind 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 0.200 m. What is the speed of the bullet?

Explanation / Answer

Let the speed of bullet be v m/s

Let us consider the inelastic collison
Use conservation of momentum
m1*v1i = (m1+m2)*Vf
9.2*10^-3 * v = (9.2*10^-3 + 2.43)* Vf
Vf = 3.772*10^-3 v

Now use conservation of energy
0.5*(m1+m2)*Vf^2 = 0.5*K*A^2
(m1+m2)*Vf^2 = K*A^2
(9.2*10^-3 + 2.43)* ( 3.772*10^-3 v)^2 = 857 * (0.2)^2
(9.2*10^-3 + 2.43)* ( 3.772*10^-3 v)^2 = 857 * (0.2)^2
( 3.772*10^-3 v)^2 = 14.054
3.772*10^-3 * v = 3.749
v = 1007.2 m/s
Answer: 1007.2 m/s

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