Question

A bullet of mass 0.01 kg is fired horizontally into 4kg wooden block at rest on a horizontal surface. the coefficient of kinetic friction between the block and the surface is 0.25. The bullet remains embedded in the block and the combination moves 20m before coming to rest. with what speed did the bullet strike the block.

Answer 1

mass of bullet, m = 0.01 kg

mass of wooden block, M = 4 kg

initial velocity of wooden block, u = 0

Let v be the velocity of system after bullet has entered into the block. the system comes to rest after covering a distance, s = 20m. Let a is the retardation produced by the system due to the frictional force, then

v² = u² + 2as

v² = 0 + 2 × 20 × a

v² = 40a.......(1)

the kinetic frictional force must be equal to retardation force.

e.g., (m + M)a = $\mu$(m + M)g

or, a = $\mu g$

a = 0.25 × 10 = 2.5 m/s² , put it in equation (1),

v² = 40 × 2.5 = 100

=> v = 10 m/s

now use law of conservation of linear momentum for collision of bullet with the block.

so, $mv_b=(m+M)v$

or, $v_b=\frac{(m+M)v}{m}$

= (0.01 + 4) × 10/0.01

= (4.01) × 1000

= 4010 m/s

hence, velocity of bullet is 4010 m/s

Answer 2

Answer:

your Answer is given above.