A block of wood of mass 0.91 kg is initially at rest. A bullet of mass 0.047 kg traveling at 104.7 m/s strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?
Use the conservation of momentum.
Massbullet*veloctiybullet=(combinedmass)newveloctiy
5.14m/s
is my answer correct
To find the speed at which the block of wood and the bullet move just after the collision, we need to apply the law of conservation of momentum.
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision in an isolated system.
Before the collision, the block of wood is at rest, so its initial momentum is zero (since momentum, p = mass x velocity, and the velocity is zero).
Let's assume the final velocity of the block of wood and the bullet after the collision is v. Since the bullet becomes embedded in the block, their final velocities would be the same.
The total initial momentum before the collision is given by:
Initial momentum = bullet's momentum before the collision = (mass of bullet) x (velocity of bullet)
Total initial momentum = (0.047 kg) x (104.7 m/s)
According to the law of conservation of momentum, the total final momentum after the collision is equal to the total initial momentum.
Total final momentum = Total initial momentum
(mass of block + mass of bullet) x final velocity = (mass of bullet) x (velocity of bullet)
(0.91 kg + 0.047 kg) x v = (0.047 kg) x (104.7 m/s)
Now, we can solve for the final velocity, v:
(0.957 kg) x v = (0.047 kg) x (104.7 m/s)
v = [(0.047 kg) x (104.7 m/s)] / (0.957 kg)
Calculating this, we find:
v ≈ 5.11 m/s
Therefore, the block of wood and the bullet move at a speed of approximately 5.11 m/s just after the collision.