A bullet with mass 200 g hits a 4.80 kg block horizontally with a speed of 400 m

ID: 1509376 • Letter: A

Question

A bullet with mass 200 g hits a 4.80 kg block horizontally with a speed of 400 m/s.

The horizontal track is 28 m long and has a kinetic coefficient of friction uk = 0.4.

a) what would be the block's velocity at point C where the spring's length is equal to it's natural length?

b) How much does the block compress the spring (k = 4.00 kN/m) before momentarily coming to rest?

c) Where does the block ultimitely come to rest?

d) What would be the frequency of spring oscillation if we assume the block sticks to the spring and oscillates on the frictionless surface?

Vo-400m/s B 28m

momentum conservation for bullet and block (for bullet final velocity = 0, for block initial velocity = 0)

m* u = mass of block * v

v= 0.2 * 400/ 4.8 = 16 m/s ( velocity with which block moves)

F = ma

mu * mg = ma

a = mu* g

v2 - u2 = 2as

v2 - (16*16) = 2 * 0.4 * 9.8 * 28

a) v = 21.8 m/s

conservation of energy for block and spring.

1/2 m v2 = 1/2 k x2

compression in spring x = 4.8 * 21.8*21.8/ 4* 103 = 0.75 m

c) Block come to rest at a distace = 28.75 m

d) frequency f = (1/ 2* pi ) sqrt(m/k)

= 5.51 Hz

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