A block of wood of mass 1.2kg initially at rest is free to slide on a horizontal surface. A bullet of mass 100g travelling at a speed of 780ms^-1 is fired horizontally at the block and becomes embedded in it. Calculate the speed the block and bullet move off at immediately after the collision.
conservation of momentum
massbullet*velocitybullet=(massbullet+massblock)V
solve for V
v=1.67
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant.
Step 1: Calculate the initial momentum of the bullet before the collision.
The initial momentum of the bullet can be calculated as the product of its mass and velocity:
Initial momentum of the bullet = mass of the bullet x velocity of the bullet
= 0.1 kg x 780 m/s
= 78 kg m/s
Step 2: Calculate the initial momentum of the block before the collision.
Since the block is initially at rest, its initial momentum is zero:
Initial momentum of the block = 0 kg m/s
Step 3: Calculate the total initial momentum before the collision.
The total initial momentum is the sum of the initial momenta of the bullet and the block:
Total initial momentum = Initial momentum of the bullet + Initial momentum of the block
= 78 kg m/s + 0 kg m/s
= 78 kg m/s
Step 4: Calculate the final momentum of the combined block and bullet system after the collision.
According to the principle of conservation of momentum, the total momentum after the collision is equal to the total momentum before the collision.
Final momentum of the combined system = Total initial momentum
= 78 kg m/s
Step 5: Calculate the final velocity of the block and bullet after the collision.
The final velocity can be calculated by dividing the final momentum by the combined mass of the block and bullet:
Final velocity = Final momentum / (mass of the bullet + mass of the block)
= 78 kg m/s / (1.2 kg + 0.1 kg)
≈ 62.5 m/s
Therefore, the speed at which the block and bullet move off immediately after the collision is approximately 62.5 m/s.