- #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 ?