Question

A $25$ gram bullet $(0.025kg)$ is fired from a gun with a speed of $210m{s^{ - 1}}$ If the gun has a mass of $0.91kg$ when rounded to two significant figures what is the recoil speed of the gun?

Answer 1

Hint: In order to solve this question, we will first find the total initial momentum of the system (bullet and gun) and then total final momentum of the system and by using principle of linear conservation of momentum and putting values of given known parameters we will solve for recoil speed of the gun up to two significant figures.

Formula used:
If P and P’ be the initial and final momentum of a system then, according to the law of conservation of momentum we have, $P = P'$ where momentum is simply the product between mass m and velocity v of a body $p = mv$.

Complete step by step solution:
According to the question, we have given that before firing the bullet from the gun the whole system was at rest so initial momentum let’s say P, it will be zero.
$P = 0 \to (i)$
Now, we have ${m_{bullet}} = 0.025kg$ mass of the bullet
${v_{bullet}} = 210m{s^{ - 1}}$ speed of the bullet
So, momentum of bullet let’s say $P{'_{bullet}}$ will be $P{'_{bullet}} = 0.025 \times 210 \to (A)$
and, ${m_{gun}} = 0.91kg$ mass of the gun
let’s say v denotes for velocity of the gun then momentum of gun will be
$P{'_{gun}} = 0.91 \times v \to (B)$
so, total final momentum of the system let’s say P’ will be the sum of equations (A) and (B) so we get,
$P' = 0.025 \times 210 + 0.91 \times v \to (ii)$
Now, using law of conservation of linear momentum and equating equations (i) and (ii) we get,
$0 = 0.025 \times 210 + 0.91 \times v$
on solving for v we get,
$v = - (\dfrac{{0.025 \times 210}}{{0.91}})$
$ \Rightarrow v = - 5.769m{s^{ - 1}}$
on rounding off for two significant figures we get,
$v = - 5.8m{s^{ - 1}}$ the negative sign shows that the gun moves backward as compared to the direction of the bullet. Hence, the recoil speed of the gun is $5.8m{s^{ - 1}}$.

Note:
It should be remembered that, mass and velocity should be in their respective SI units while calculating such equations and the basic rule of rounding off is that if a digit is greater than $5$ we just add $1$ to the digit prior to the digit up to where we are rounding off any value such as here on two places after decimal we have a value of $6( > 5)$ in $ - 5.769m{s^{ - 1}}$ so we simply add $1$ to the $7$ and its rounded off to the value of $5.8m{s^{ - 1}}.$