A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the ceiling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take g=98msâˆ’2

Question

A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the ceiling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take g=98msâˆ’2

Answer 1

http://www.jiskha.com/display.cgi?id=1279532750 as an example.

Answer 2

To determine the speed of the bullet before it collided with the block, we can use the principle of conservation of momentum and the conservation of mechanical energy.

1. Conservation of Momentum:
Before the collision, the bullet is moving horizontally, and the block is at rest. The total momentum before the collision is zero. After the collision, the bullet becomes lodged in the block, and they move together.

Let 'm1' be the mass of the bullet and 'm2' be the mass of the block. Let 'v' be the speed of the bullet before the collision. After the collision, the bullet and the block together have a mass of (m1 + m2).

Using conservation of momentum:
(m1 * v) + (0 * 0) = (m1 + m2) * final_velocity

Since the bullet and the block move together after the collision, the final velocity is the same for both, which we'll call 'vf'. The equation becomes:
(m1 * v) = (m1 + m2) * vf

2. Conservation of Mechanical Energy:
The bullet and the block rise to a height of 0.30 meters above the original height. We can calculate the change in gravitational potential energy.

The change in gravitational potential energy is equal to the work done by the gravitational force:
∆PE = mgh

Here, 'm' represents the total mass of the bullet and block, 'g' represents the acceleration due to gravity (98 m/s²), and 'h' represents the change in height (0.30 m).

3. Putting it all together:
Now we have two equations:
(m1 * v) = (m1 + m2) * vf (Equation 1)
∆PE = (m1 + m2) * g * h (Equation 2)

Rearrange Equation 1 to solve for 'v':
v = (m1 + m2) * vf / m1

Plug this value of 'v' into Equation 2 and solve for 'vf':
∆PE = (m1 + m2) * g * h
(m1 + m2) * vf / m1 = (m1 + m2) * g * h
vf = (g * h * m1) / (m1 + m2)

Plugging in the given values:
g = 98 m/s²
h = 0.3 m
m1 = 10 g = 0.01 kg
m2 = 2 kg

vf = (98 * 0.3 * 0.01) / (0.01 + 2) ≈ 0.147 m/s

Therefore, the speed of the bullet before the collision was approximately 0.147 m/s.

http://www.jiskha.com/display.cgi?id=1279532750 as an example.

133,72

To determine the speed of the bullet before it collided with the block, we can use the principle of conservation of momentum and the conservation of mechanical energy.