A 8.38g bullet is fired into a 4.60kg block suspended as in the figure. The bullet stops in the block, which rises 15.0cm above its initial position. Find initial speed of the bullet.
To find the initial speed of the bullet, we can use the principle of conservation of linear momentum.
The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant if no external forces are acting on it.
In this case, before the bullet is fired into the block, the system is isolated in the horizontal direction. Therefore, the total linear momentum before the collision is equal to the total linear momentum after the collision.
The total linear momentum before the collision is given by the formula:
(m1 * v1) + (m2 * v2) = (m1 + m2) * V
where
m1 = mass of the bullet (8.38g = 0.00838kg)
v1 = initial velocity of the bullet (what we're trying to find)
m2 = mass of the block (4.60kg)
v2 = initial velocity of the block (which is initially at rest, so v2 = 0)
V = final velocity of the combined bullet-block system after the collision
Since the bullet stops in the block and the block rises, we can assume that the final velocity V is upwards and can be calculated using the concept of gravitational potential energy.
The gravitational potential energy gained by an object when it rises to a certain height is given by the formula:
PE = m * g * h
where
m = mass of the object (the combined mass of the bullet and block = 4.60kg + 0.00838kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = vertical displacement (15.0cm = 0.15m)
Setting the gravitational potential energy equal to the kinetic energy of the bullet-block system after the collision:
(PE) = (KE)
(m * g * h) = (1/2) * (m1 + m2) * V^2
Now we can solve for V.
Substituting the values, we have:
(4.60838kg * 9.8 m/s^2 * 0.15m) = (1/2) * (4.60kg + 0.00838kg) * V^2
Simplifying the equation gives:
0.67835676 = (2.30417kg) * V^2
Dividing both sides of the equation by (2.30417kg) gives:
V^2 = 0.2942
Taking the square root of both sides gives:
V = 0.542 m/s
Now that we have the final velocity of the bullet-block system, we can find the initial velocity of the bullet (v1) by rearranging the initial momentum equation:
(m1 * v1) + (m2 * v2) = (m1 + m2) * V
Since v2 = 0, the equation simplifies to:
m1 * v1 = (m1 + m2) * V
Substituting the values, we have:
0.00838kg * v1 = (0.00838kg + 4.60kg) * 0.542 m/s
Simplifying the equation gives:
0.00838kg * v1 = 2.52156 kg*m/s
Dividing both sides of the equation by 0.00838kg gives:
v1 = 301.07 m/s
Therefore, the initial speed of the bullet is approximately 301.07 m/s.