Solveeit Logo
Login

Question

Question: A \(25\) gram bullet \((0.025kg)\) is fired from a gun with a speed of \(210m{s^{ - 1}}\) If the gun...

A 2525 gram bullet (0.025kg)(0.025kg) is fired from a gun with a speed of 210ms−1210m{s^{ - 1}} If the gun has a mass of 0.91kg0.91kg when rounded to two significant figures what is the recoil speed of the gun?

Explanation

Solution

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′P = P' where momentum is simply the product between mass m and velocity v of a body p=mvp = 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→(i)P = 0 \to (i)
Now, we have mbullet=0.025kg{m_{bullet}} = 0.025kg mass of the bullet
vbullet=210ms−1{v_{bullet}} = 210m{s^{ - 1}} speed of the bullet
So, momentum of bullet let’s say Pbullet′P{'_{bullet}} will be Pbullet′=0.025×210→(A)P{'_{bullet}} = 0.025 \times 210 \to (A)
and, mgun=0.91kg{m_{gun}} = 0.91kg mass of the gun
let’s say v denotes for velocity of the gun then momentum of gun will be
Pgun′=0.91×v→(B)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′=0.025×210+0.91×v→(ii)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=0.025×210+0.91×v0 = 0.025 \times 210 + 0.91 \times v
on solving for v we get,
v=−(0.025×2100.91)v = - (\dfrac{{0.025 \times 210}}{{0.91}})
⇒v=−5.769ms−1\Rightarrow v = - 5.769m{s^{ - 1}}
on rounding off for two significant figures we get,
v=−5.8ms−1v = - 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 5.8ms−15.8m{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 55 we just add 11 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)6( > 5) in −5.769ms−1 - 5.769m{s^{ - 1}} so we simply add 11 to the 77 and its rounded off to the value of 5.8ms−1.5.8m{s^{ - 1}}.