The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.030 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.055 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.
(a) Find the final speed of puck A.
(b) Find the final speed of puck B.
i got part b which was 2.8 but have no idea how to get part a alitle help please. thank you
remember, conservation of momentum is a vector, so do it in two directions, x, and y. Conservation of energy is just one equation. You can get three equations out of these.
To find the final speed of puck A after the collision, you need to apply the principles of conservation of momentum and conservation of kinetic energy.
(a) Find the final speed of puck A:
1. Determine the initial momentum of puck A:
Initial momentum = mass of puck A * initial velocity of puck A
Initial momentum = 0.030 kg * 5.5 m/s = 0.165 kg·m/s
2. Determine the initial momentum of puck B:
Since puck B is initially at rest, its initial momentum is zero.
3. Determine the total initial momentum:
Total initial momentum = initial momentum of puck A + initial momentum of puck B
Total initial momentum = 0.165 kg·m/s + 0 kg·m/s = 0.165 kg·m/s
4. Determine the final momentum of puck A:
Since momentum is conserved in a collision, the final momentum of puck A will be equal to the total initial momentum.
5. Determine the final velocity of puck A using the equation:
Final momentum of puck A = mass of puck A * final velocity of puck A
final velocity of puck A = final momentum of puck A / mass of puck A
6. Calculate the final velocity of puck A:
final velocity of puck A = 0.165 kg·m/s / 0.030 kg ≈ 5.50 m/s
Therefore, the final speed of puck A after the collision is approximately 5.50 m/s.
(b) The final speed of puck B, as you mentioned, is 2.8 m/s.