Two speeding lead bullets each of mass 5 g and temperature of 20°C collide head on at speeds of 500m/s each. Assuming a perfectly inelastic collision and no loss of energy to the atmospheres, describe the final state of the two bullet system.
To describe the final state of the two-bullet system after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the bullets are initially at rest (speed = 0 m/s), the total initial momentum is zero.
2. Conservation of kinetic energy:
In a perfectly inelastic collision, the kinetic energy is not conserved. It is converted into other forms of energy, such as heat or deformation of the objects involved.
Now, let's calculate the final velocity of the two-bullet system after the collision:
1. Calculate the total initial momentum:
The momentum (p) of an object is given by the product of its mass (m) and velocity (v). Since both bullets have the same mass (5g or 0.005kg) and velocity (500 m/s), the total initial momentum is:
P_initial = (mass1 * velocity1) + (mass2 * velocity2)
= (0.005 kg * 500 m/s) + (0.005 kg * -500 m/s) [Note: negative sign denotes opposite direction]
= 2.5 kg·m/s - 2.5 kg·m/s
= 0 kg·m/s
2. Calculate the final velocity:
Since the total initial momentum is zero, the total final momentum must also be zero. As the bullets collide and stick together, they will move as a single system after the collision. Let's assume the final velocity of the system is V_final.
Total final momentum = (system mass) * (V_final)
0 kg·m/s = (0.01 kg) * V_final
Therefore, V_final = 0 m/s.
The final state of the two-bullet system after the collision is that it comes to a complete stop. The bullets stick together and move as a single mass at rest, with no kinetic energy left.
To describe the final state of the two bullet system after the collision, we need to apply the principles of conservation of linear momentum and conservation of kinetic energy.
Step 1: Calculate the initial momentum of each bullet.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v). Therefore, the initial momentum of each bullet is:
p = m * v
Given:
Mass of each bullet (m) = 5 g (convert to kg: 5 g ÷ 1000 = 0.005 kg)
Velocity of each bullet (v) = 500 m/s
So, the initial momentum of each bullet is:
p = 0.005 kg * 500 m/s = 2.5 kg*m/s
Step 2: Apply conservation of linear momentum.
In an isolated system, the total momentum before the collision is equal to the total momentum after the collision.
In this case, as the bullets are colliding head-on, the total initial momentum is the sum of the momenta of the two bullets before the collision.
Total initial momentum = p1 + p2 = 2.5 kg*m/s + 2.5 kg*m/s = 5 kg*m/s.
In a perfectly inelastic collision, the two objects stick together after the collision. Therefore, the final mass (mf) will be the sum of the individual masses of the two bullets:
Final mass (mf) = m1 + m2 = 0.005 kg + 0.005 kg = 0.01 kg.
Step 3: Apply conservation of kinetic energy.
In a perfectly inelastic collision, kinetic energy is not conserved. The initial kinetic energy (KEi) of the system is given by the sum of the kinetic energies of the two bullets before the collision.
Initial kinetic energy = (1/2) m1 * v1^2 + (1/2) m2 * v2^2
Given:
Mass of each bullet (m) = 5 g (0.005 kg)
Velocity of each bullet (v) = 500 m/s
Initial kinetic energy = (1/2) * 0.005 kg * (500 m/s)^2 + (1/2) * 0.005 kg * (500 m/s)^2
= 1.25 * 10^5 J
Since energy is not conserved in the collision, the final kinetic energy (KEf) will not remain the same.
Final kinetic energy = KEf = 0 (as it is a perfectly inelastic collision).
Step 4: Solve for the final velocity.
Using the conservation of linear momentum and the fact that the final velocity is the same for both bullets:
Total final momentum = (mf) * (vf)
Total initial momentum = (mi) * (vi)
Where:
mi = initial mass of each bullet = m
vi = initial velocity of each bullet = v (same for both bullets)
mf = final mass of the combined system
vf = final velocity of the combined system
Substituting the values:
5 kg*m/s = (0.01 kg) * (vf)
vf = 5 kg*m/s / 0.01 kg
vf = 500 m/s
So, the final velocity of the two-bullet system after the collision is 500 m/s.
Therefore, the final state of the two-bullet system is a single object with a mass of 0.01 kg, moving at a velocity of 500 m/s.
The final state? Are you assuming no heat was lost
heat= total KE= 2*.06*500^2 * 1/2 joules
temperatue rise: above KE heat= .1*specificheatcapacityLead*deltaTemp
solve for deltatemmp