A 0.250-kg ice puck, moving east with a speed of 5.26m/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 0.250-kg puck after the collision?
1m/s
To find the speed of the 0.250-kg puck after the collision, we need to apply the principles of conservation of momentum.
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.
Let's denote the initial velocity of the 0.250-kg puck as v1i, and the final velocity as v1f. Similarly, let's denote the initial velocity of the 0.990-kg puck as v2i.
Using the conservation of momentum, we can write the equation:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
Where:
m1 = mass of the 0.250-kg puck = 0.250 kg
m2 = mass of the 0.990-kg puck = 0.990 kg
v1i = initial velocity of the 0.250-kg puck = 5.26 m/s
v2i = initial velocity of the 0.990-kg puck = 0 m/s (as it is initially at rest)
v1f = final velocity of the 0.250-kg puck (what we want to find)
v2f = final velocity of the 0.990-kg puck (0.990-kg puck is initially at rest, so its final velocity will be v1i)
Now we can plug in the values into the equation:
(0.250 kg) * (5.26 m/s) + (0.990 kg) * (0 m/s) = (0.250 kg) * v1f + (0.990 kg) * (5.26 m/s)
Working out the equation:
(1.315 kg•m/s) = (0.250 kg) * v1f + (5.2034 kg•m/s)
Rearranging to solve for v1f:
v1f = (1.315 kg•m/s - 5.2034 kg•m/s) / (0.250 kg)
v1f = -3.8884 kg•m/s / 0.250 kg
v1f = -15.5536 m/s
The negative sign indicates that the 0.250-kg puck is moving in the opposite direction. So, the speed of the 0.250-kg puck after the collision is approximately 15.55 m/s in the west direction.