A block of mass M = 5.4 kg, at rest on horizontal frictionless table, is attache

Question

A block of mass M = 5.4 kg, at rest on horizontal frictionless table, is attached to a rigid support by a spring of constant, k = 7400N/m. A bullet of mass m = 10g and a velocity v of magnitude 1000m/s strikes and is embedded in the block. Assuming the compression of the spring is negligible until the bullet is embedded, determine the a) speed of the block immediately after the collision, b) the kinetic energy of the block immediately after the collision, c) the maximum potential energy of the block after the collision and d) amplitude of the resulting simple harmonic motion.

Explanation / Answer

Given that

M = 5.4kg, m = 10gms = 0.01kg

V = 1000m/s

k = 7400N/m

a) The speed immediately after the collision is

Va = (m/m+M)V

Va = (0.01*1000)/(5.4+0.01)

Va = 10/5.41 = 1.848m/s

Va = 1.848m/s

b & c) The energy of the spring-object system is constant since there are no external work done on the system (no fricition), therefore

1/ 2 (m + M )va^ 2 = 1/ 2 kxmax^2

K.E = P.E

K.E = (0.01+5.4)1.848^2/2

K.E = 5.41*3.415104/2

K.E = 9.2378 J

d) Amplitude A = [1/(m+M)k]^1/2*mV

A = [1/(5.41*7400)]^1/2 *0.01*1000

A = (1/40034)^1/2 *10

A = (2.497*10^-5)^1/2*10

A = 0.01064*10

A = 0.1064m

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