A bullet A of mass 2kg travelling at 200m/s emblezes itself in a wooden block of mass 9kg moving in the opposite direction. At 20m/s. Calculate the velocity of the block when the bullet comes to rest inside the block. And what principle is used in the direction
Initial momentum
= 2(200) -9(20) = 400 - 180 = 220
final momentum = 11 v
Conservation of momentum, Newton's First Law --->
220 = 11 v
so
v = 20 m/s
To solve this problem, we can use the principle of conservation of linear momentum. According to this principle, the total linear momentum before the collision is equal to the total linear momentum after the collision.
The linear momentum (p) of an object is given by the product of its mass (m) and velocity (v):
p = m * v
Before the collision:
The bullet has a mass of 2 kg and is traveling at 200 m/s, so its momentum is:
p_bullet_before = 2 kg * 200 m/s = 400 kg·m/s
The wooden block has a mass of 9 kg and is moving in the opposite direction at 20 m/s, so its momentum is:
p_block_before = 9 kg * (-20 m/s) = -180 kg·m/s (negative because it's moving in the opposite direction)
Total momentum before the collision:
p_total_before = p_bullet_before + p_block_before
= 400 kg·m/s - 180 kg·m/s
= 220 kg·m/s
After the collision:
The bullet comes to rest inside the block, which means their final velocity will be the same.
Let's assume the final velocity of the bullet and the block after the collision is v_final. Since the bullet comes to rest, its final linear momentum is zero:
p_bullet_after = 2 kg * v_final = 0 kg·m/s
The block's final linear momentum can be calculated as:
p_block_after = 9 kg * v_final
Total momentum after the collision:
p_total_after = p_bullet_after + p_block_after
= 0 kg·m/s + 9 kg * v_final
= 9 kg * v_final
Using the conservation of linear momentum principle, we can equate the total momentum before the collision to the total momentum after the collision:
p_total_before = p_total_after
220 kg·m/s = 9 kg * v_final
To find the velocity (v_final) of the block after the collision, we can rearrange the equation:
v_final = 220 kg·m/s / 9 kg
≈ 24.44 m/s
Therefore, the velocity of the wooden block after the bullet comes to rest inside it is approximately 24.44 m/s.