Question

A bullet of mass 0.1 kg is fired from a gun weighing 5.0 kg. If the initial speed of the bullet is 250 m/s, calculate the speed with which the gun recoils.

a) 5 m/s
b) 0.5 m/s
c) 0.05 m/s
d) 50 m/s

Answer 1

The speed with which the gun recoils is 5 m/s in the opposite direction.

Step-by-step explanation:

To find the speed with which the gun recoils, we can use the law of conservation of momentum. According to this law, the total momentum before the bullet is fired is equal to the total momentum after the bullet is fired. The momentum of an object is given by the product of its mass and velocity.

Let the initial velocity of the gun be V. The mass of the bullet is 0.1 kg and its initial velocity is 250 m/s. The mass of the gun is 5.0 kg and its recoil velocity is V', which we need to find.

Using the conservation of momentum, we can write:

(0.1 kg * 250 m/s) + (5.0 kg * V) = 0

0.1 kg * 250 m/s = -5.0 kg * V' (Since the bullet moves in the opposite direction of the gun recoil)

Simplifying the equation, we get:

V' = (-0.1 kg * 250 m/s) / 5.0 kg = -5 m/s

Therefore, the speed with which the gun recoils is 5 m/s in the opposite direction.

Answer 2

`5\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Apply the conservation of momentum to solve this question.

When an object of mass
`m` travels at a velocity of
`v`, the momentum
`p` of this object would be:

`\begin{aligned}p = m\, v\end{aligned}`.

Let
`m_(1) = 0.1\; {\rm kg}` and
`v_(1) = 250\; {\rm m\cdot s^(-1)}` denote the mass and new velocity of the bullet. Let
`m_(2) = 5.0\; {\rm kg}` and
`v_(2)` (to be found) denote the mass and new velocity of the gun.

`(\text{total final momentum}) = m_(1)\, v_(1) + m_(2)\, v_(2)`.

Assuming that there was initially no motion, such that total initial momentum is zero:

`(\text{total initial momentum}) = 0`.

By the conservation of momentum:

`(\text{total final momentum}) = (\text{total initial momentum})`.

Hence:

`m_(1)\, v_(1) + m_(2)\, v_(2) = 0`.

Rearrange this equation to find
`v_(2)`:

`\begin{aligned}v_(2) &= (-m_(1)\, v_(1))/(m_(2)) \\ &= \frac{-(0.1\; {\rm kg})\, (250\; {\rm m\cdot s^(-1)})}{5.0\; {\rm kg}} \\ &= (-5.0)\; {\rm m\cdot s^(-1)}\end{aligned}`.

(Negative because the recoil velocity points backwards.)

Speed is equal to the magnitude of velocity. Since the velocity of recoil is
`(-5.0)\; {\rm m\cdot s^(-1)}`, the speed of recoil would be
`5.0\; {\rm m\cdot s^(-1)}`.