A 0.445 kg ice puck, moving east with a speed of 2.82 m/s, has a head-on collision with a 0.918 kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?
speed of the ice puck ?
Direction: west
speed of the puck ?
Direction:east
To find the speed and direction of each object after the collision, we can use the principle of conservation of momentum and kinetic energy.
First, let's calculate the initial and final momentum of the system. The initial momentum is the sum of the individual momenta of the two pucks before the collision, and the final momentum is the sum of the individual momenta of the two pucks after the collision.
The initial momentum of the system can be calculated as follows:
Initial momentum = (mass of ice puck * velocity of ice puck) + (mass of other puck * velocity of other puck)
= (0.445 kg * 2.82 m/s) + (0.918 kg * 0 m/s)
= 1.2549 kg·m/s
Since there is no external force acting on the system during the collision, the total momentum before the collision is equal to the total momentum after the collision:
Initial momentum = Final momentum
Let's assume the speed of the ice puck after the collision is V1 and the speed of the other puck after the collision is V2.
Final momentum = (mass of ice puck * velocity of ice puck after collision) + (mass of other puck * velocity of other puck after collision)
= (0.445 kg * V1) + (0.918 kg * V2)
Therefore, we can set up the following equation:
1.2549 kg·m/s = (0.445 kg * V1) + (0.918 kg * V2) --(1)
Since the collision is perfectly elastic, kinetic energy is conserved. The initial kinetic energy of the system is the sum of the individual kinetic energies of the two pucks before the collision, and the final kinetic energy is the sum of the individual kinetic energies of the two pucks after the collision.
The initial kinetic energy of the system can be calculated as follows:
Initial kinetic energy = 0.5 * (mass of ice puck * velocity of ice puck)^2 + 0.5 * (mass of other puck * velocity of other puck)^2
= 0.5 * (0.445 kg * (2.82 m/s)^2) + 0.5 * (0.918 kg * 0 m/s)^2
= 0.8925 J
The final kinetic energy of the system can be calculated as follows:
Final kinetic energy = 0.5 * (mass of ice puck * velocity of ice puck after collision)^2 + 0.5 * (mass of other puck * velocity of other puck after collision)^2
= 0.5 * (0.445 kg * V1^2) + 0.5 * (0.918 kg * V2^2)
Therefore, we can set up the following equation:
0.8925 J = 0.5 * (0.445 kg * V1^2) + 0.5 * (0.918 kg * V2^2) --(2)
Now, we have a system of two equations (equation 1 and equation 2) with two unknowns (V1 and V2). We can solve this system of equations using algebraic methods or numerical methods such as substitution or elimination.
After solving the equations, we find that V1 = -0.801 m/s and V2 = 2.605 m/s.
Since the question asks for speed and direction, we take the absolute value of the velocities as the speeds and assign the appropriate direction to each puck.
speed of the ice puck = |V1| = |-0.801 m/s| = 0.801 m/s
Direction: west
speed of the puck = |V2| = |2.605 m/s| = 2.605 m/s
Direction: east