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

Question

A block of mass M = 5.4 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 6000 N/m. A bullet of mass m = 9.5 g and velocity V of magnitude 630 m/s strikes and is embedded in the block (Fig. 15-39). Assuming the compression of the spring is negligible until the bullet is embedded, determine the speed of the block immediately after the collision and the amplitude of the resulting simple harmonic motion. In Fig. 15-40, block 2 of mass 2.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos (omega t + pi/2). Block 1 of

Explanation / Answer

use conservation of mometum,

initial momentum = final momentum

=> M*0 + m* 630m/s= ( M+ m)* v

=> v= 1.081 m/s

now kinetic energy at this point will be potential energy at max compression of spring

the distance of copression will be the amplitude

so ,

0.5 ( M+m)*V2= 0.5k*x2

=>

x=0.033 m

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