a bullet of mass 0.01kg and travelling at a speed of 500m/s strikes a block of mass 2kg which is suspended by a string length 5m. the centre of gravity of block is found to rise a vertical distance of 0.1m .Calculate the speed of the bullet when it emerges from the block.

220m/s

To calculate the speed of the bullet when it emerges from the block, we can use the principle of conservation of momentum and the principle of conservation of mechanical energy.

Let's assume the speed of the bullet after emerging from the block is v, and the final velocity of the block and the bullet together after the collision is V.

1. Conservation of momentum:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Initial momentum before the collision = mass of the bullet × velocity of the bullet
Final momentum after the collision = (mass of the bullet + mass of the block) × velocity after the collision (V)

0.01 kg × 500 m/s = (0.01 kg + 2 kg) × V
5 kg m/s = 2.01 kg × V
V = 5 kg m/s / 2.01 kg
V ≈ 2.49 m/s

2. Conservation of mechanical energy:
The change in gravitational potential energy of the block is related to the mechanical energy lost due to the collision.

Gravitational potential energy change = mass of the block × acceleration due to gravity × change in height
Potential energy change = 2 kg × 9.8 m/s^2 × 0.1 m
Potential energy change = 1.96 J

The mechanical energy lost is equal to the change in potential energy:
Mechanical energy lost = potential energy change = 1.96 J

The initial kinetic energy of the bullet is equal to the sum of the mechanical energy lost and the final kinetic energy of the block and bullet after the collision:

Initial kinetic energy = mechanical energy lost + final kinetic energy
0.5 × 0.01 kg × (500 m/s)^2 = 1.96 J + 0.5 × (2.01 kg) × (2.49 m/s)^2
12.5 J = 1.96 J + 2.475 J
12.5 J - 1.96 J = 2.475 J
10.54 J = 2.475 J

The final kinetic energy of the bullet after emerging from the block is 10.54 J.

3. Using the final kinetic energy:
The final kinetic energy can be calculated using the equation:

Final kinetic energy = 0.5 × mass of the bullet × (velocity of the bullet after emerging from the block)^2

10.54 J = 0.5 × 0.01 kg × (velocity of the bullet after emerging from the block)^2
10.54 J = 0.005 kg × (velocity of the bullet after emerging from the block)^2

Solving for the velocity of the bullet after emerging from the block:

(velocity of the bullet after emerging from the block)^2 = 10.54 J / 0.005 kg
(velocity of the bullet after emerging from the block)^2 = 2108 J/kg

Taking the square root on both sides:

velocity of the bullet after emerging from the block = √(2108 J/kg)
velocity of the bullet after emerging from the block ≈ 45.95 m/s

Therefore, the speed of the bullet when it emerges from the block is approximately 45.95 m/s.

To calculate the speed of the bullet when it emerges from the block, we can apply the law of conservation of linear momentum and the principles of work and energy.

1. Determine the initial momentum of the bullet:
The linear momentum (p) of an object is given by the formula p = mass × velocity.
Given that the mass of the bullet (m1) is 0.01 kg and its initial velocity (v1) is 500 m/s, the initial momentum (p1) of the bullet is:
p1 = m1 × v1 = 0.01 kg × 500 m/s = 5 kg·m/s.

2. Determine the final momentum of the bullet-block system:
Since the bullet is embedded in the block, the final momentum (pf) of the system will be the sum of the momenta of the bullet and the block.
However, before we can calculate the final momentum, we need to determine the velocity of the block-bullet system right after the impact.

3. Derive the velocity of the block-bullet system after the impact:
When the bullet strikes the block, it becomes embedded in it. This forms a combined mass (m2) of the bullet and the block, i.e., m2 = m1 + m_block.
After the impact, the velocity of the block-bullet system (v2) can be determined using the principle of conservation of linear momentum:
p1 = p_block + p_bullet
m1 × v1 = m_block × v2
(m1 + m_block) × v2 = m1 × v1
(0.01 kg + 2 kg) × v2 = 0.01 kg × 500 m/s
(2.01 kg) × v2 = 5 kg·m/s
v2 = 5 kg·m/s / 2.01 kg
v2 ≈ 2.49 m/s

4. Calculate the change in potential energy:
The center of gravity of the block rises vertically a distance of 0.1 m, which implies that work is done against gravity.
The change in potential energy (ΔPE) can be calculated using the formula ΔPE = m_block × g × h, where m_block is the mass of the block, g is the acceleration due to gravity (9.8 m/s²), and h is the vertical distance the block rises.
ΔPE = 2 kg × 9.8 m/s² × 0.1 m
ΔPE = 1.96 kg·m²/s²

5. Derive the final velocity of the bullet after the block emerges:
According to the work-energy theorem, the work done on the bullet during the collision is equal to the change in its kinetic energy (ΔKE).
ΔKE = ΔPE
1/2 × m_bullet × v_final² - 1/2 × m_bullet × v_initial² = 1.96 kg·m²/s²
1/2 × 0.01 kg × v_final² - 1/2 × 0.01 kg × 500 m/s² = 1.96 kg·m²/s²
0.005 kg × v_final² - 2500 kg·m²/s² = 1.96 kg·m²/s²
0.005 kg × v_final² = 1.96 kg·m²/s² + 2500 kg·m²/s²
0.005 kg × v_final² = 2501.96 kg·m²/s²
v_final² = (2501.96 kg·m²/s²) / (0.005 kg)
v_final ≈ √500,392 m²/s²
v_final ≈ 707.11 m/s

Therefore, the speed of the bullet when it emerges from the block is approximately 707.11 m/s.

energy of block in motion:

1/2 MV^2=Mgh
V= sqrt(2g*.1)
momentum of bullet and block after bullet leaves: MV+mv=m(500) (convervation momentum)
solve for v
2(sqrt(.2*9.8))+.01*v=(.01)(500)
v= (5-2*1.4)/.01 = you are correct. 220m/s