A bullet of mass m is shot at speed v toward a pendulum bob of mass M. The bullet penetrates through the block and emerges on the other side traveling at speed v/2.
1. What is the speed v_block of the block immediately after the bullet emerges from the block (In terms of M, m and v)?
2. To what maximum height h does the block rise (In terms of M, m and v)?
m-mass of bullet, v- initial velocity of bullet, let v'be the velocity of block.
conservation of momentum,
m*v+ M*0= m*(v/2)+ M*v'
solve for v'.
To answer these questions, we can use the principles of conservation of momentum and conservation of energy. Let's break it down step by step:
1. Calculation of the speed of the block (v_block):
- Conservation of momentum: In the absence of external forces, the total momentum before the collision is equal to the total momentum after the collision.
- Before the collision: The bullet has mass m and speed v, and the block has mass M and speed 0.
- After the collision: The bullet has mass m and speed v/2, and the block has mass M and speed v_block.
- Using the conservation of momentum, we can write the equation: m * v + 0 = m * (v/2) + M * v_block.
- Solving the equation for v_block, we get: v_block = (2m*v - m*(v/2)) / M.
- Simplifying, we have: v_block = (2mv - mv) / (2M) = mv / (2M).
Therefore, the speed of the block immediately after the bullet emerges from the block is v_block = mv / (2M).
2. Calculation of the maximum height reached by the block (h):
- Conservation of energy: The total energy before the collision is equal to the total energy after the collision.
- Before the collision: The bullet has kinetic energy given by (1/2) * m * v^2, and the block has gravitational potential energy equal to M * g * h (where g is the acceleration due to gravity).
- After the collision: The bullet has kinetic energy given by (1/2) * m * (v/2)^2, and the block has gravitational potential energy equal to M * g * h.
- Using the conservation of energy, we can write the equation: (1/2) * m * v^2 + M * g * h = (1/2) * m * (v/2)^2 + M * g * h.
- Simplifying and rearranging the equation, we get: (1/2) * m * v^2 - (1/2) * m * (v/2)^2 = M * g * h.
- Simplifying further, we have: [(1/2) * m * v^2] - [(1/2) * m * (v^2)/4] = M * g * h.
- Combining like terms, we get: [(4/4) - (1/4)] * (1/2) * m * v^2 = M * g * h.
- Simplifying and solving the equation for h, we have: (3/4) * (1/2) * m * v^2 = M * g * h.
- Simplifying further, we get: (3/8) * m * v^2 = M * g * h.
- Solving the equation for h, we finally have: h = (3/8) * (m/M) * v^2 / g.
Therefore, the maximum height reached by the block is h = (3/8) * (m/M) * v^2 / g.
To answer your questions, we can apply the principle of conservation of momentum and the principle of conservation of energy.
1. What is the speed v_block of the block immediately after the bullet emerges from the block (In terms of M, m, and v)?
According to the principle of conservation of momentum, the total momentum before and after the collision should be the same.
Before the collision:
Momentum of the bullet = m * v
Momentum of the block = 0 (since it's initially at rest)
After the collision:
Momentum of the bullet = m * (v / 2) (since it emerges at half the speed)
Momentum of the block = v_block * M
Applying the conservation of momentum, we can write:
m * v = m * (v / 2) + v_block * M
Simplifying the equation:
2 * m * v = m * v + v_block * M
2 * m * v - m * v = v_block * M
v_block = m * v / M
Therefore, the speed of the block immediately after the bullet emerges from it is v_block = (m * v) / M.
2. To what maximum height h does the block rise (In terms of M, m, and v)?
To determine the maximum height the block rises, we can apply the principle of conservation of mechanical energy.
The initial mechanical energy of the system is equal to the final mechanical energy.
Initial mechanical energy:
- Initial kinetic energy of the bullet: (1/2) * m * v^2
- Initial potential energy of the block: 0 (since it starts from rest)
Final mechanical energy:
- Final kinetic energy of the bullet: (1/2) * m * (v/2)^2 (since it emerges with half the speed)
- Final potential energy of the block: M * g * h (where g is the acceleration due to gravity and h is the maximum height)
Equating the initial and final mechanical energies, we can write:
(1/2) * m * v^2 + 0 = (1/2) * m * (v/2)^2 + M * g * h
Simplifying the equation:
(1/2) * m * v^2 = (1/2) * m * (v/2)^2 + M * g * h
m * v^2 = (1/4) * m * v^2 + M * g * h
3/4 * m * v^2 = M * g * h
Solving for h, the maximum height:
h = (3/4 * m * v^2) / (M * g)
Therefore, the maximum height the block rises is h = (3/4 * m * v^2) / (M * g).
velocity imparted to block will cause the the block to rise to a height h at this instant the kinetic energy is converted into potential energy completely so,
(1/2)*M*(v')^2 = M*g*h
use the value of v' from above solution and solve for h.