Question

a bullet of mass 20gram moving with a velocity of 300m/s gets embedded in a freely suspended wooden block of mass 880 garm what is the velocity acquired by the block ?​

Answer 1

Given :

▪ Mass of bullet = 20g

▪ Mass of block = 880g

▪ Initial velocity of bullet = 300m/s

To Find :

▪ Velocity of system (block + bullet) after collision.

Concept :

↗ This question is completely based on the concept of momentum conservation.

↗ Here, no net force acts on the system so we can apply this concept.

Calculation :

$\rightrightarrows\tt\:P_i=P_f\\ \\ \rightrightarrows\tt\:(P_i)_{bullet}+(P_i)_{block}=(P_f)_{system}\\ \\ \rightrightarrows\tt\:(m\times v)+(M\times V)=(M+m)V'\\ \\ \rightrightarrows\tt\:(0.02\times 300)+(0.88\times 0)=(0.02+0.88)V'\\ \\ \rightrightarrows\tt\:6+0=0.9V'\\ \\ \rightrightarrows\underline{\boxed{\bf{\red{V'=6.67\:mps}}}}\:\orange{\bigstar}$

Answer 2

Explanation:

This is completely inelastic collision. The linear momentum is conserved.

If a body of mass, m collides with a body of mass M and gets embedded in it, according to law of conservation of momentum,

mv =( m+M)V.

Here,

m is mass of colliding body. m= 20 g.

M is mass of target body. M=880 g.

v=300m/s is velocity of colliding body.

V= ? V is velocity of combined system of m and M.

Therefore 20(g).300(m/s)=(880+20)( g)V(m/s.)OR

V=6000/900=6.67 m/s.