- #1
ac7597
- 126
- 6
- Homework Statement
- A hockey puck of mass m=0.36 kg slides in the positive X direction at v=13.8 m/s. It collides with a stationary brick of mass M=1.35 kg.
After the collision, each object slides away to the right. The puck slides at a new speed u=10.410 m/s, at an angle of α above the X-axis, while the brick slides at a speed w=1.340 m/s at a different angle β below the X-axis.
This collision is : inelastic
What is the angle α?
- Relevant Equations
- momentum=(mass)(velocity)
x(before) | y(before) | x(after) | y(after) | |
puck | (0.36)(13.8)=4.968 | 0 | (0.36)(10.41)cos(α) | (0.36)(10.41)sin(α) |
brick | 0 | 0 | (1.35)(1.34)cos(β) | -(1.35)(1.34)sin(β) |
total | 4.968 N*s | 0 | 4.968 N*s | 0 |
thus: (0.36)(10.41)sin(α) =(1.35)(1.34)sin(β)
β= sin^(-1)[3.747sin(α)/1.809]= sin^(-1)[2.071sin(α)]
(0.36)(10.41)cos(α) + (1.35)(1.34)cos(β) = 4.968
3.747cos(α) + 1.809cos(sin^(-1)[2.071sin(α)] ) =4.968
3.747cos(α) + 1.809 (1-(2.071sin(α))^(2) )^(1/2) = 4.968
2.071cos(α) + (1-(2.071sin(α))^ (2) )^(1/2) = 2.746
4.29cos^(2)(α) + (1-(2.071sin(α))^(2) ) = 7.54
4.29cos^(2)(α) + 1-4.29sin^2(α) = 7.54
4.29cos^(2)(α) -4.29(1-cos^2(α)) = 6.54
4.29cos^(2)(α) -4.29 + 4.29cos^2(α) = 6.54
8.58cos^(2)(α) = 10.83
α=undefined ?