A bullet whose mass is 0.0015kg leaves a 5kg leaves a 5kg rifle at a muzzle velocity of 800 m/s. It strikes a 7kg block of wood and becomes embedded in it.
After the bullet hits the rock, what is the speed of the bullet/ block system?
preserve momentum, so
(0.0015)(800) = (0.0015+7)v
To determine the speed of the bullet/block system after the bullet hits the block, we can use the law of conservation of momentum. According to this law, the total momentum before the collision should be equal to the total momentum after the collision.
Before the collision, the bullet's momentum can be calculated using the formula: momentum = mass × velocity. The mass of the bullet is given as 0.0015 kg, and the velocity is given as 800 m/s. Thus, the bullet's momentum before the collision is:
Momentum of bullet before collision = (mass of bullet) × (velocity of bullet)
= 0.0015 kg × 800 m/s
After the collision, the bullet becomes embedded in the block of wood, so the momentum of the bullet/block system will be equal to the final momentum after the collision. Let's assume the velocity of the bullet/block system after the collision is V (in m/s).
Momentum of bullet/block system after collision = (mass of bullet + mass of block) × (velocity of bullet/block system)
Given that the mass of the bullet is 0.0015 kg and the mass of the block is 7 kg, we can write the equation for momentum conservation:
(initial momentum before collision) = (final momentum after collision)
0.0015 kg × 800 m/s = (0.0015 kg + 7 kg) × V
Now, we can solve for V:
V = (0.0015 kg × 800 m/s) / (0.0015 kg + 7 kg)
By plugging in the values and performing the calculation, we can find the speed of the bullet/block system after the collision.