A bullet of mass 0.102 kg traveling horizontally at a speed of 300 m/s embeds it

Question

A bullet of mass 0.102 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3.5 kg that is sitting on a nearly frictionless surface? (a) What is the speed of the block after the bullet embeds itself in the block? (b) Calculate the kinetic energy of the bullet plus the block before the collision? (c) Calculate the kinetic energy of the bullet plus the block after the collision? (d) Was this collision elastic or inelastic? (e) Calculate the rise in internal energy of the bullet plus block system?

MY QUESTION IS......

What is the choice of system to answer each question? Is it bullet and block?

Explanation / Answer

Choice of system????

You have to consider both the elements of the system i.e., bullet and the block.

I mean to say, the bullet has an initial speed but the block has not.

But finally both the elements have speed.

So, use conservation of momentum.

(a) By the law of momentum conservation:-
=>m1u1+m2u2=(m1+m2)v
=>0.102 x 300 = (0.102+3.5) x v
=>v = 8.49 m/s
(b) KE(before) = 1/2 x m1 x u1^2 = 1/2 x 0.102 x (300)^2 = 4590 J
(c) KE(after) = 1/2 x (m1+m2) x v^2 = 1/2 x 3.602 x (8.49)^2 = 129.82 J

(d) Since initial kinetic energy is not equal to final kinetic energy so the collision is inelastic.

(e) I think this question should be rise in thermal energy because final kinetic energy is less which is ultimately converting into heat.

So, rise in thermal energy = Loss in KE = KE(before) - KE(after) = 4590 - 129.82 = 4460.18 J

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