a bullet of mass 0.1kg moving with velocity 500 m/s strikes with a block of mass 2kg suspended with a string.after collision block is raised by 0.1m. what will be the velocity of bullet after the collision. bullet is not embeded in the block.
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To determine the velocity of the bullet after the collision, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and its velocity (v): p = m * v.
Before the collision:
The momentum of the bullet before the collision is given by: p_bullet_before = m_bullet * v_bullet_before.
The momentum of the block before the collision is given by: p_block_before = m_block * v_block_before, where v_block_before is zero since the block is suspended.
After the collision:
The momentum of the bullet after the collision is given by: p_bullet_after = m_bullet * v_bullet_after.
The momentum of the block after the collision is still zero, as the block is suspended in equilibrium.
Since the block is lifted after the collision, the bullet must provide an upward impulse to the block, causing it to rise. The impulse (I) is equal to the change in momentum and is given by: I = p_bullet_after - p_bullet_before.
Therefore, we can rewrite the impulse equation as: I = (m_bullet * v_bullet_after) - (m_bullet * v_bullet_before).
We know that the block is raised by 0.1 m, which means the work done is equal to the change in potential energy:
Work = m_block * g * h, where g is the acceleration due to gravity and h is the height.
Since the force applied to the block is in the upward direction, and the displacement is also in the upward direction, the work done is positive.
Equating work done to the impulse, we have: I = m_block * g * h.
Now we can substitute the values given in the problem:
m_bullet = 0.1 kg (mass of the bullet)
v_bullet_before = 500 m/s (velocity of the bullet before the collision)
m_block = 2 kg (mass of the block)
h = 0.1 m (height the block is raised)
Now we can equate the impulse and the work done equation: (m_bullet * v_bullet_after) - (m_bullet * v_bullet_before) = m_block * g * h.
Rearranging the equation, we can solve for v_bullet_after: v_bullet_after = ((m_bullet * v_bullet_before) + (m_block * g * h)) / m_bullet.
Plugging in the values, we get: v_bullet_after = ((0.1 kg * 500 m/s) + (2 kg * 9.8 m/s^2 * 0.1 m)) / 0.1 kg.
Calculating the value, the velocity of the bullet after the collision is approximately 521 m/s.