A bullet of mass 12.4 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.02 kg, is subject to no horizontal external forces during the collision with the bullet. After the collision, the block is observed to move at a speed of 4.80 m/s.
(a) Find the initial speed of the bullet.
(b) How much kinetic energy is lost?
The law of conservation of linear momentum for inelastic collision
m•v =(m +M) •u
v =(m +M) •u/m =(0.0124+1.02) •4.8/0.0124 = 399.6 m/s
KE1 =m•v^2/2 =0.0124•(399.6)^2/2 =990 J
KE2 = (m+M) •u^2/2=11.9 J
KE1 –KE2 =990 -11.9 = 978.1J
To solve this problem, we can apply the Principle of Conservation of Momentum and the Principle of Conservation of Kinetic Energy.
(a) To find the initial speed of the bullet, we can start by using the Principle of Conservation of Momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision. The momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v).
Before the collision:
The bullet has a mass of 12.4 g, which is 0.0124 kg. Let the initial velocity of the bullet be v1.
The block is initially stationary, so its velocity is 0 kg/s.
After the collision:
The bullet comes to rest in the block, so its velocity becomes 0 m/s.
The block is observed to move at a speed of 4.80 m/s.
Applying the Principle of Conservation of Momentum:
Total momentum before = total momentum after
(0.0124 kg)(v1) + (1.02 kg)(0 m/s) = (0.0124 kg)(0 m/s) + (1.02 kg)(4.80 m/s)
Simplifying the equation:
0.0124 kg v1 = 1.02 kg (4.80 m/s)
Solving for v1:
v1 = (1.02 kg)(4.80 m/s) / 0.0124 kg
v1 = 394.19 m/s
Therefore, the initial speed of the bullet is approximately 394.19 m/s.
(b) To find how much kinetic energy is lost during the collision, we can use the Principle of Conservation of Kinetic Energy. The initial kinetic energy is given by the formula KE = 0.5mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
Before the collision:
The bullet has an initial mass of 0.0124 kg and an initial velocity of v1.
After the collision:
The bullet comes to rest, so its final velocity is 0 m/s.
The block has a final velocity of 4.80 m/s.
Applying the Principle of Conservation of Kinetic Energy:
Initial kinetic energy = Final kinetic energy
(0.5)(0.0124 kg)(v1)^2 = (0.5)(0 kg)(0 m/s)^2 + (0.5)(1.02 kg)(4.80 m/s)^2
Simplifying the equation:
0.0062 kg (v1)^2 = 0 + 11.71 J
Solving for (v1)^2:
(v1)^2 = 11.71 J / 0.0062 kg
(v1)^2 = 1888.71 m^2/s^2
Taking the square root to find v1:
v1 = √(1888.71 m^2/s^2)
v1 = 43.44 m/s
The initial speed of the bullet is approximately 43.44 m/s.
To calculate the amount of kinetic energy lost, we can subtract the final kinetic energy from the initial kinetic energy:
Initial kinetic energy - Final kinetic energy
(0.5)(0.0124 kg)(43.44 m/s)^2 - (0.5)(1.02 kg)(4.80 m/s)^2 = 31.64 J
Therefore, approximately 31.64 Joules of kinetic energy is lost during the collision.