Question
Question: A bullet of mass \(25\,g\) is fired with velocity \(400\,m{s^{ - 1}}\) at a bag of sand of mass \(4....
A bullet of mass 25g is fired with velocity 400ms−1 at a bag of sand of mass 4.975kg suspended by a rope and gets embedded into it. What is the velocity gained by the bag?
A) 0.2ms−1
B) 4ms−1
C) 0.4ms−1
D) 2ms−1
Solution
Hint:- In this problem, the mass of the bullet is given and the velocity of the bullet is given and the mass of the sandbag is also given, then the velocity gained by the bag can be determined by the conserving the momentum of the system.
Useful formula:
Momentum of an object is equal to the product of mass and velocity,
p=m×v
Where, p is the momentum of the object, m is the mass of the object and v is the velocity of the object
Complete step by step solution:
Given that,
Mass of bullet, mb=25g,
Velocity of the bullet, vb=400ms−1,
The mass of the sandbag, ms=4.975kg.
The momentum of the bullet is equal to the momentum of the sandbag, then
mb×vb=(mb+ms)×vs...............(1)
In RHS the mass of the bullet is added with the mass of the sand because after firing the bullet is put into the sand bag, so the mass of the bullet is added with the mass of the sandbag.
Now substituting the mass of the bullet, velocity of the bullet and the mass of the sandbag in the equation (1), then
0.025×400=(0.025+4.975)×vs
On multiplying the terms in LHS and adding the terms in RHS, then
10=5×vs
By keeping the term vs in one side and the other terms in others side, then
vs=510
On dividing the above equation, then
vs=2ms−1
Thus, the above equation shows the velocity of the sandbag.
Hence, the option (D) is correct.
Note: The mass of the bullet is added with the mass of the sand bag because after firing the bullet, the bullet from the gun hits the sandbag and gets inside the sandbag, so the mass of the bullet is also considered for the momentum of the sandbag.