a bullet of mass0.01 kg is fired into a sandbag of mass 0.49kg hanging from a tree. the sandbag with the bullet embedded into it, swings away at 10m/s. find momentum after
mv = (m+M)u
p= (m+M)u =(0.01+0.49)•10= 5 kg•m/s
habtshtade
a,5Kgm/s
To find the momentum after the bullet is embedded into the sandbag, we can use the principle of conservation of momentum.
The equation for momentum is given by:
Momentum (p) = mass (m) * velocity (v)
Where:
- The momentum is measured in kg*m/s
- The mass is measured in kg
- The velocity is measured in m/s
In the given scenario, the bullet is fired into the sandbag, so the initial momentum is equal to the momentum of the bullet before it hits the sandbag.
The initial momentum of the bullet is calculated as follows:
Initial momentum of bullet = mass of bullet * velocity of bullet
Initial momentum of bullet = 0.01 kg * (velocity of bullet) --- (1)
Also, the sandbag with the bullet embedded in it swings away at 10 m/s. Let's call this final velocity v'.
So, the final momentum is given by:
Final momentum of bullet and sandbag = (mass of bullet + mass of sandbag) * final velocity
Final momentum of bullet and sandbag = (0.01 kg + 0.49 kg) * 10 m/s
Final momentum of bullet and sandbag = 0.5 kg * 10 m/s = 5 kg*m/s --- (2)
According to the principle of conservation of momentum, the initial momentum should be equal to the final momentum:
(1) = (2)
0.01 kg * (velocity of bullet) = 5 kg*m/s
To find the velocity of the bullet, we can rearrange the equation:
velocity of bullet = 5 kg*m/s / 0.01 kg
velocity of bullet = 500 m/s
Therefore, the momentum after the bullet is embedded into the sandbag is 5 kg*m/s.