A 2.00 g bullet hits and becomes embedded in a 5.00-kg woodblock which is hangin

Question

A 2.00 g bullet hits and becomes embedded in a 5.00-kg woodblock which is hanging from a 1.20 m long string. This causesthe block to swing through an arc of 3.50 degrees. What wasthe speed of the bullet before it hit the block? The final answer is 524 m/s but don't know how to getthat. A 2.00 g bullet hits and becomes embedded in a 5.00-kg woodblock which is hanging from a 1.20 m long string. This causesthe block to swing through an arc of 3.50 degrees. What wasthe speed of the bullet before it hit the block? The final answer is 524 m/s but don't know how to getthat.

Explanation / Answer

mass of bullet m = 2 g = 0.002 kg mass of vblock M = 5 kg Length of the string L = 1.2 m angle = 3.5 degree Total mass M ' = m + M                       = 5.002 kg Kinetic energy just after collision = potential energy atmaximum height      ( 1/ 2) M ' v ^ 2 = M ' gh                          v = [2gh ] height h = L ( 1- cos )              =2.238* 10 ^ -3 m So, v = 0.2094 m / s from law of conservation of momentum , m u + M U = M ' v where U = initial speed of the block = 0 So, mu = M ' v Therefore initial speed of the bullet u = M ' v /m                                                         = 524 m / s from this   h = v ^ 2 / 2g

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