A bullet of mass 0.05kg is fired horizontally into 10kg block which is free to move if both bullet and block move with common velocity find the velocity with which the bullet hit the block and
conserve momentum.
If the bullet hit with velocity v1
and the combination moved with velocity v2, then
0.05 v1 + 10*0 = 10.05 v2
maybe if you finish your question things can go a bit further ...
To find the velocity with which the bullet hits the block, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the velocity of the bullet before the collision as v_b and the velocity of the block before the collision as v_block. Since the bullet is fired horizontally, its initial vertical velocity is zero.
The momentum of an object is calculated by multiplying its mass by its velocity. The total momentum before the collision is given by:
Total momentum before collision = (mass of bullet) x (velocity of bullet) + (mass of block) x (velocity of block)
Given that the mass of the bullet is 0.05 kg and the mass of the block is 10 kg, the equation becomes:
Total momentum before collision = (0.05 kg) x (v_b) + (10 kg) x (v_block)
After the collision, the bullet and the block move with the same velocity, which we'll denote as v_common. This means that the final velocity of the bullet and the block is the same.
The total momentum after the collision is given by:
Total momentum after collision = (0.05 kg) x (v_common) + (10 kg) x (v_common)
According to the principle of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. Therefore, we can set these two equations equal to each other:
(0.05 kg) x (v_b) + (10 kg) x (v_block) = (0.05 kg) x (v_common) + (10 kg) x (v_common)
Simplifying this equation, we have:
0.05 kg x v_b + 10 kg x v_block = 0.05 kg x v_common + 10 kg x v_common
Now we have an equation that relates the velocities of the bullet (v_b), block (v_block), and the common velocity (v_common).
Since the question states that the bullet and block move with a common velocity, we can set the velocities of the bullet and the block as equal to the common velocity:
v_b = v_common
v_block = v_common
Substituting these values into the equation, we get:
0.05 kg x v_common + 10 kg x v_common = 0.05 kg x v_common + 10 kg x v_common
Simplifying this equation, we have:
0.05 kg x v_common + 10 kg x v_common = 0.05 kg x v_common + 10 kg x v_common
Now we can solve for v_common:
0.05 kg x v_common + 10 kg x v_common = 0.05 kg x v_common + 10 kg x v_common
Combining like terms, we get:
10.05 kg x v_common = 10.05 kg x v_common
Dividing both sides by 10.05 kg, we obtain:
v_common = v_common
Therefore, the velocity with which the bullet hits the block is equal to the common velocity, which implies that the bullet and the block move together with the same velocity.