Physics Help Forum Bullet impact

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Apr 15th 2009, 12:13 PM #1 Junior Member   Join Date: Feb 2009 Posts: 8 Bullet impact A 12 g bullet is fired into a 100 g block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.65, what is the speed of the bullet immediately before impact?
 Apr 16th 2009, 10:07 AM #2 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 M = mass of the block m = mass of the bullet u = velocity of the bullet before impact U = velocity of block+bullet just after the impact. a = accelleration of the block+bullet after impact k = coefficient of kinetic friction g = accln due to gravity. x = distance travelled by block+bullet (=7.5 m) before comming to rest. The friction force = - k(M+m)g hence acceln of block+bullet = -kg $\displaystyle V^2 = U^2 + 2ax$ $\displaystyle 0 = U^2 -2kgx$ U = sqrt{2kgx} Now we use law of conservation of momentum mu = (M+m)U =(M+m)*sqrt{2kgx} u = (M+m)*sqrt{2kgx}/m

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