A 1.00 E- 2 kg bullet is Fired horizontally into a 2.50kg wooden block attached

Question

A 1.00 E- 2 kg bullet is Fired horizontally into a 2.50kg wooden block attached to the end of a massles horizontal spring, k= 845 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 strickes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision the spring is compressed a long it's axis and causes the block/ bullet to oscillate with an amplitude of 0.200m. A) is the kinetic energy of the bullet plus the block the same as the kinetic energy of the bullet? Why or why not? B) if so use kinetic energy to determine the velocity of the bullet. If not, use conservation of momentum where the momentum of the bullet (mv) equals the momentum of the bullet and block (m+M) V, a totally inelastic collision.

Explanation / Answer

Mass of bullet m = 1 x 10 -2 kg Mass of block M = 2.5 kg Spring constant k = 845 N /m Amplitude A= 0.2 m A) No, the kinetic energy of the bullet plus the block the same as the kinetic energy of the bullet. Because it is a inelastic collision. In ineleastic collision ,kinetic energy is not conserved. B) used conservation of momentum. From law of conservation of momentum , mv+MU = (m+M) V             V = mv / (m+M)    where MU = momentum of the block = 0 Since U = 0             V = 0.003984 v        -----( 1) From law of conservation of energy , K.E of the bullet block system = Elastic P.E of the spring        ( 1/ 2) (m+ M) V 2 = ( 1/ 2) kA 2 From this V = A [ k / (m+M) ]                     = 3.669 m / s 0.003984 v = 3.669 Speed of the bullet v = 921 m / s              

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