A 15 g bullet strikes and embeds in a 2.8 kg block suspended at the end of a 1 m string. After the collision the string rises to a maximum angle of 20 degrees to the vertical.
what is the speed of bullet?
figure how high the blocks/bullet went.
then figure potential energy (mgh) it increased. That is equal to the initial KE of the block/bullet
figure the velocity of the block/bullet from the KE.
Now,having the initial veloctiy block/bullet,you can find the initial momentum of the bullet from conservation of momentum,thence its velocity.
To find the speed of the bullet, we can use the principle of conservation of momentum.
The momentum before the collision is equal to the momentum after the collision.
1. Calculate the momentum before the collision:
Momentum = mass × velocity
The mass of the bullet is 15 g, which is equal to 0.015 kg.
Let's assume the initial velocity of the bullet is v.
Momentum before the collision = (mass of bullet) × (initial velocity of bullet)
Momentum before the collision = 0.015 kg × v
Momentum before the collision = 0.015v kg*m/s
2. Calculate the momentum after the collision:
The bullet embeds in the block after the collision, so the combined mass of the bullet and block is the mass of the block.
The mass of the block is 2.8 kg.
The final velocity of the bullet and block can be calculated by using the concept of conservation of momentum.
Momentum after the collision = (mass of block + bullet) × (final velocity of block and bullet)
Since the bullet is embedded in the block, the final velocity of the block and bullet is the same.
Momentum after the collision = (mass of block) × (final velocity of block and bullet)
Momentum after the collision = 2.8 kg × (final velocity of block and bullet)
Momentum after the collision = 2.8(final velocity of block and bullet) kg*m/s
3. Set the momentum before the collision equal to the momentum after the collision:
0.015v = 2.8(final velocity of block and bullet)
4. Rearrange the equation to solve for v:
v = 2.8(final velocity of block and bullet)/0.015
5. To find the final velocity of the block and bullet, we can use the conservation of mechanical energy. The change in potential energy from the initial position to the maximum angle is equal to the change in kinetic energy.
ΔPE = ΔKE
ΔPE = m × g × h
ΔPE = (mass of block) × (acceleration due to gravity) × (height)
ΔKE = 0.5 × m × v^2
ΔKE = 0.5 × (mass of block + bullet) × v^2
Since the height in this case is the length of the string, which is given as 1 m, we can substitute the values and solve for the final velocity.
(mass of block) × (acceleration due to gravity) × (height) = 0.5 × (mass of block + bullet) × v^2
2.8 kg × 9.8 m/s^2 × 1 m = 0.5 × (2.8 kg + 0.015 kg) × v^2
6. Simplify the equation and solve for v:
2.8 × 9.8 = 0.5 × (2.815 kg) × v^2
27.44 = 1.4075 × v^2
v^2 = 27.44 / 1.4075
v^2 ≈ 19.5014
v ≈ √19.5014
v ≈ 4.417 m/s
Therefore, the speed of the bullet is approximately 4.417 m/s.
To calculate the speed of the bullet, we can use the principle of conservation of energy.
Step 1: Determine the potential energy of the system before and after the collision.
- Before the collision, the system only has potential energy due to the height of the block. The potential energy (PE) is given by PE = mgh, where m is the mass of the block and h is the height.
- The mass of the block, m = 2.8 kg
- The gravitational acceleration, g = 9.8 m/s²
- The height, h = 1 m
- After the collision, the block rises to a maximum angle with the vertical. At this point, all the potential energy is converted into kinetic energy.
- The angle, θ = 20 degrees = 0.35 radians (approximately)
Step 2: Set up the equation using the conservation of energy principle.
- The initial potential energy (PE_initial) is equal to the final kinetic energy (KE_final).
- PE_initial = KE_final
- mgh = 1/2 mv²
- Cancelling out the mass, we get gh = 1/2 v²
Step 3: Solve for the velocity (v).
- Rearrange the equation to isolate v: v² = 2gh
- Take the square root of both sides: v = √(2gh)
- Substitute the known values: v = √(2 * 9.8 m/s² * 1 m) = √(19.6) ≈ 4.427 m/s
Therefore, the speed of the bullet is approximately 4.43 m/s.