A 20 g bullet traveling at 300 m/s is shot into a 4.0 kg block at rest on a fric

Question

A 20 g bullet traveling at 300 m/s is shot into a 4.0 kg block at rest on a frictionless horizontal surface. The bullet remains lodged intothe block

a) What is the speed of the block right after shooting?

b) The block then slams into an ideal spring moving horizontally ona frictionless floor. The spring constant of the spring is 2400 N/m. What is the maximum compression of the spring?

c)If the floor is rough and has a coefficent of kinetic friction of -0.45, what is the maximum compression of the spring?

Explanation / Answer

initial momentum = 0.02 * 300 = 6

final momentum = 4.01 * v

conserving momentum

intial = final

so ... 4.01 * v = 6

so v = 1.49626 m/sec

a) speed of block after shooting = 1.49626 m/sec

b)

kinetic energy of block = 0.5 * 4.01 * 1.49626^2 = 4.488781945138 J

This energy will be converted to potential energy of spring

so

0.5* 2400 * x^2 = 4.488781945138

so compression x = 0.061161 m = 6.1161 cm

c)

Let the compression be x ..

workdone by friction = 0.45 * 4.01 * 9.8 *x = 17.6841 * x

so pot energy of spring = 4.488781945138 - work done by friction

0.5 * 2400 * x^2 = 4.488781945138 - 17.6841 * x

1200x^2 + 17.6841 * x - 4.488781945138 = 0

on solving we get

x = 0.05423474644 m and x = -0.06897149644

as x cannot be negative ..

so compression x = 0.05423474644 m = 5.423474644 cm

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