A bullet of mass 4.7 g is fired horizontally into a 2.4 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.28. The bullet stops in the block, which slides straight ahead for 2.2 m (without rotation). (a) What is the speed of the block immediately after the bullet stops relative to it? (b) At what speed is the bullet fired?
work done by block/bullet= totalmass*g*mu*distance
from that, compute the initial velocity of the block/bullet system (workdone=1/2 mv^2)
Now, knowing that velocity, use the conservation of momentum to find the initial velocity of the bullet
massbullet*vbullet= totalmass*Vyoufound
To solve this problem, we need to understand the principles of conservation of momentum and kinetic energy.
(a) What is the speed of the block immediately after the bullet stops relative to it?
To find the speed of the block after the bullet stops, we first need to calculate the initial momentum of the system before the bullet hits the block. Let's denote the initial velocity of the bullet as V_bullet and the velocity of the block as V_block. Since the bullet is fired horizontally, its initial vertical velocity is zero.
The momentum of the bullet and the block before the collision can be calculated as follows:
Momentum_bullet = mass_bullet * V_bullet
Momentum_block = mass_block * V_block
After the bullet stops, the momentum of the system is conserved. Therefore, the total momentum after the collision is zero:
Momentum_total = 0 = Momentum_bullet + Momentum_block
Since the bullet stops, we can solve the equation for the velocity of the block:
0 = mass_bullet * V_bullet + mass_block * V_block
Solving for V_block, we get:
V_block = - (mass_bullet / mass_block) * V_bullet
We interpreted the negative sign to indicate that the velocity of the block is in the opposite direction to the velocity of the bullet.
(b) At what speed is the bullet fired?
To find the speed at which the bullet is fired, we can use the principle of conservation of kinetic energy. Before the collision, only the bullet possesses kinetic energy. After the collision, all the kinetic energy is transferred to the block. Therefore, the initial kinetic energy of the bullet is equal to the final kinetic energy of the block.
The kinetic energy of the bullet before the collision can be calculated as follows:
Kinetic_energy_bullet = (1/2) * mass_bullet * V_bullet^2
The kinetic energy of the block after the collision can be calculated as follows:
Kinetic_energy_block = (1/2) * mass_block * V_block^2
Since the bullet stops and the block slides forward, we have:
Kinetic_energy_block = (1/2) * mass_block * V_block^2 = (1/2) * mass_block * V_block^2
Setting these two equations equal to each other, we can solve for V_bullet:
(1/2) * mass_bullet * V_bullet^2 = (1/2) * mass_block * V_block^2
V_bullet^2 = (mass_block / mass_bullet) * V_block^2
Taking the square root of both sides, we find:
V_bullet = sqrt((mass_block / mass_bullet) * V_block^2)
Substituting the previously calculated value of V_block, we can find V_bullet.