A 0.300-kg ice puck, moving east with a speed of 5.86m/s , has a head-on collision with a 0.990-kg puck initially at rest. Assume that the collision is perfectly elastic.
What is the speed of the 0.300-kg puck after the collision?
What ise the speed of the 0.990-kg puck after the collision?
To find the speed of each puck after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
In an isolated system, the momentum before the collision is equal to the momentum after the collision. The momentum is calculated by multiplying mass and velocity.
Let's denote the initial velocity of the 0.300-kg puck as v1 initial, and the initial velocity of the 0.990-kg puck as v2 initial.
Momentum before collision (p1 initial) = m1 * v1 initial
Momentum before collision (p2 initial) = m2 * v2 initial
After the collision, the 0.990-kg puck is at rest, so its velocity (v2 final) is zero. We need to find the velocity of the 0.300-kg puck (v1 final).
The total momentum after the collision is zero, as there is no net external force acting on the system:
p1 final + p2 final = 0
So, after rearranging the equation, we get:
m1 * v1 final + m2 * v2 final = 0
(0.300 kg) * (v1 final) + (0.990 kg) * 0 = 0
Solving for v1 final, we find that the velocity of the 0.300-kg puck after the collision is zero.
2. Conservation of kinetic energy:
In an elastic collision, kinetic energy is also conserved. The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Kinetic energy before collision (KE1 initial) = 1/2 * m1 * (v1 initial)^2
Kinetic energy before collision (KE2 initial) = 1/2 * m2 * (v2 initial)^2
Kinetic energy after collision (KE1 final) = 1/2 * m1 * (v1 final)^2
Kinetic energy after collision (KE2 final) = 1/2 * m2 * (v2 final)^2
Since the 0.990-kg puck comes to rest after the collision (v2 final = 0), its kinetic energy after the collision is zero.
So, applying the conservation of kinetic energy, we have:
KE1 initial + KE2 initial = KE1 final + KE2 final
1/2 * (0.300 kg) * (5.86 m/s)^2 + 0 = 1/2 * (0.300 kg) * (v1 final)^2 + 0
Simplifying the equation, we find:
1/2 * (0.300 kg) * (5.86 m/s)^2 = 1/2 * (0.300 kg) * (v1 final)^2
Now we can solve for v1 final by rearranging and calculating the square root:
(v1 final)^2 = (0.300 kg) * (5.86 m/s)^2 / (0.300 kg)
(v1 final)^2 = (5.86 m/s)^2
v1 final = 5.86 m/s
Therefore, the speed of the 0.300-kg puck after the collision will be 5.86 m/s. And as already mentioned, the speed of the 0.990-kg puck will be zero.