Question

A bullet of mass 250 gram strikes horizontally to a wooden block of mass 4750 gram at a speed of 100 metre per second if the friction between the floor and wooden block is 100 Newton find the distance after which the system will come?​

Answer 1

Answer:

s = 62.5 cm

Explanation:

Mass of the bullet,  $\sf{m=0.25\ kg}$

Mass of the block,  $\sf{M=4.75\ kg}$

Initial velocity of the bullet,  $\sf{u = 100\ ms^{-1}}$

Since the block was in rest, its initial velocity,  $\sf{U=0\ ms^{-1}}$

After striking, the bullet and the block combine together and move with the same velocity.

By law of conservation of linear momentum,

$\sf{mu+MU=mv+Mv}\\\\\\\sf{(0.25\times100)+(4.75\times0)}=(0.25+4.75)v\\\\\\\sf{5v=25}\\\\\\\sf{v=5\ ms^{-1}}$

So the combined system move with this velocity after striking.

Consider the combined system after striking. The initial velocity $\sf{=v=5\ ms^{-1}}$

The system will come to rest after moving some distance which has to be found. Then, final velocity $\sf{=0\ ms^{-1}}$

Retardation is provided by the frictional force, i.e.,

$\sf{(m+M)a=-100\ N}\\\\\\\sf{5a=-100}\\\\\\\sf{a=-20\ ms^{-2}}$

Frictional force has negative sign because it acts opposite to the displacement.

Now, by the third kinematic equation,

$\sf{v^2=u^2+2as,}$

we have,

$\sf{0^2=5^2+2\times-20s}\\\\\\\sf{25-40s=0}\\\\\\\sf{\Large\underline{\underline{s=0.625\ m}}}$

So the system will come to rest after travelling 62.5 cm.