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 determine the speed of each puck after the collision, we need to apply the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
The total momentum of a system is conserved before and after the collision. Mathematically, this can be expressed as:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
where m1 and m2 are the masses of the two pucks, v1i and v2i are their initial velocities, and v1f and v2f are their final velocities.
2. Conservation of kinetic energy:
In a perfectly elastic collision, the kinetic energy of the system is conserved. Mathematically, this can be expressed as:
(1/2) * m1 * v1i^2 + (1/2) * m2 * v2i^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2
Let's plug in the given values and solve for the final velocities:
m1 = 0.300 kg (mass of the first puck)
v1i = 5.86 m/s (initial velocity of the first puck)
m2 = 0.990 kg (mass of the second puck)
v2i = 0 m/s (initial velocity of the second puck, as it is initially at rest)
Using the conservation of momentum equation:
0.300 kg * 5.86 m/s + 0.990 kg * 0 m/s = 0.300 kg * v1f + 0.990 kg * v2f
Simplifying the equation:
1.758 kg·m/s = 0.300 kg * v1f + 0 kg·m/s (since the second puck is initially at rest)
Therefore, we have:
0.300 kg * v1f = 1.758 kg·m/s
v1f = 1.758 kg·m/s / 0.300 kg
v1f ≈ 5.86 m/s
So, the speed of the 0.300-kg puck after the collision is approximately 5.86 m/s.
To find the speed of the 0.990-kg puck after the collision, we need to use the fact that the total momentum of the system is conserved:
0.300 kg * 5.86 m/s + 0.990 kg * 0 m/s = 0.300 kg * 5.86 m/s + 0.990 kg * v2f
Again, simplifying the equation:
1.758 kg·m/s = 0.300 kg * 5.86 m/s + 0.990 kg * v2f
Solving for v2f:
0.990 kg * v2f = 1.758 kg·m/s - 0.300 kg * 5.86 m/s
v2f ≈ (1.758 kg·m/s - 0.300 kg * 5.86 m/s) / 0.990 kg
Calculating v2f:
v2f ≈ 1.758 kg·m/s / 0.990 kg
v2f ≈ 1.776 m/s
Therefore, the speed of the 0.990-kg puck after the collision is approximately 1.776 m/s.