A 0.400-kg ice puck, moving east with a speed of 5.38 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
What is the speed of the 0.400-kg puck after the collision?
To solve this problem, we can use the law of conservation of momentum.
The law of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.
Let's denote the velocity of the 0.400-kg puck after the collision as v1, and the velocity of the 0.950-kg puck after the collision as v2.
Before the collision, the momentum of the system is the sum of the momentum of the two pucks:
p_initial = m1 * v1_initial + m2 * v2_initial
After the collision, the momentum of the system is:
p_final = m1 * v1 + m2 * v2
Since the collision is perfectly elastic, the total kinetic energy of the system is conserved. This means that the sum of the kinetic energies before the collision is equal to the sum of the kinetic energies after the collision:
(1/2) * m1 * (v1_initial)^2 + (1/2) * m2 * (v2_initial)^2 = (1/2) * m1 * (v1)^2 + (1/2) * m2 * (v2)^2
Now we can solve these equations to find the velocity of the 0.400-kg puck after the collision, v1.
First, let's calculate the initial momentum of the system:
p_initial = (0.400 kg * 5.38 m/s) + (0.950 kg * 0 m/s)
= 2.152 kg⋅m/s
Next, let's calculate the final momentum of the system after the collision:
p_final = (0.400 kg * v1) + (0.950 kg * v2)
Since the collision is perfectly elastic, the total momentum is conserved:
p_initial = p_final
2.152 kg⋅m/s = (0.400 kg * v1) + (0.950 kg * v2)
Now let's calculate the kinetic energies before and after the collision:
Initial kinetic energy = (1/2) * 0.400 kg * (5.38 m/s)^2 + (1/2) * 0.950 kg * (0 m/s)^2
= 5.765 J
Final kinetic energy = (1/2) * 0.400 kg * (v1)^2 + (1/2) * 0.950 kg * (v2)^2
Since the kinetic energy is conserved, we can set the initial kinetic energy equal to the final kinetic energy:
5.765 J = (1/2) * 0.400 kg * (v1)^2 + (1/2) * 0.950 kg * (v2)^2
Now we have two equations:
2.152 kg⋅m/s = (0.400 kg * v1) + (0.950 kg * v2)
5.765 J = (1/2) * 0.400 kg * (v1)^2 + (1/2) * 0.950 kg * (v2)^2
We can use these equations to solve for v1.
To find the speed of the 0.400-kg puck after the collision, we can use 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 formula for momentum is given by:
momentum = mass × velocity
Initially, the 0.400-kg puck is moving east with a speed of 5.38 m/s, and the 0.950-kg puck is at rest. This means the initial momentum of the system is:
momentum_initial = (mass1 × velocity1) + (mass2 × velocity2)
For the 0.400-kg puck: momentum1_initial = (0.400 kg) × (5.38 m/s) = 2.152 kg·m/s
For the 0.950-kg puck: momentum2_initial = (0.950 kg) × (0 m/s) = 0 kg·m/s
Therefore, the total initial momentum is:
momentum_initial = momentum1_initial + momentum2_initial = 2.152 kg·m/s + 0 kg·m/s = 2.152 kg·m/s
Now, let's consider the situation after the collision. We are told that the collision is perfectly elastic. In an elastic collision, both momentum and kinetic energy are conserved.
Since only momentum is relevant to find the speed, we can focus on conserving momentum. After the collision, the 0.950-kg puck will be moving east, and the 0.400-kg puck will be moving west.
Let's label the final velocities as v1_final for the 0.400-kg puck, and v2_final for the 0.950-kg puck.
The final momentum can now be calculated as:
momentum_final = (mass1 × velocity1) + (mass2 × velocity2)
For the 0.400-kg puck: momentum1_final = (0.400 kg) × (-v1_final)
For the 0.950-kg puck: momentum2_final = (0.950 kg) × (v2_final)
Since the collision is elastic, the total final momentum is equal to the total initial momentum:
momentum_initial = momentum_final
2.152 kg·m/s = (0.400 kg) × (-v1_final) + (0.950 kg) × (v2_final)
We also know that the initial velocity of the 0.950-kg puck is 0 m/s, so we can rewrite the equation as:
2.152 kg·m/s = (0.400 kg) × (-v1_final) + (0.950 kg) × (0 m/s + v2_final)
Simplifying the equation gives:
2.152 kg·m/s = - 0.400 kg × v1_final + 0.950 kg × v2_final
Now, we can solve this equation for the final velocity of the 0.400-kg puck (v1_final).
To do so, we need another equation that relates the final velocities of the two pucks. In an elastic collision, the relative velocity of separation after the collision is equal to the relative velocity of approach before the collision.
Relative velocity of separation = v1_final - v2_final
Relative velocity of approach = velocity1_initial - velocity2_initial
Therefore, we have:
v1_final - v2_final = velocity1_initial - velocity2_initial
Since the 0.950-kg puck is initially at rest, we have:
v1_final - v2_final = velocity1_initial - 0 m/s
v1_final - v2_final = velocity1_initial
Plugging this relationship back into the momentum equation:
2.152 kg·m/s = - 0.400 kg × v1_final + 0.950 kg × (v1_final)
2.152 kg·m/s = (0.950 kg - 0.400 kg) × v1_final
2.152 kg·m/s = 0.550 kg × v1_final
Now, we can solve for v1_final:
v1_final = 2.152 kg·m/s / 0.550 kg
v1_final ≈ 3.91 m/s
So, the speed of the 0.400-kg puck after the collision is approximately 3.91 m/s to the west.
initial momentum east =.4*5.38
the final momentum east is the same
so
.4 * 5.38 = .4 u + .95 v
where u and v are the two resulting speeds
also kinetic energy is the same
(1/2)(.4)(5.38)^2 = (1/2)(.4)(u)^2 + (1/2)(.95)(v)^2
use those two equations to solve for u and v