A 8.34kg lump of clay travels east on frictionless ice at a speed of 2.40 m/s. A .170 kg hockey puck travels northward and collides with the clay. After the collision, both objects stick together, and move off at a speed of 2.59 m/s in a direction of 23.9* north of east. What was the hockey pucks speed before the collision.
M1*V1 + M2*V2[90o] = 2.59m/s[23.9o]
8.34*2.4 + 0.170*V2[90o] = 2.59[23.9o]
20.02 + 0.170*V2[90] = 2.37 + 1.05i
0.170V2[90] = 2.37-20.02 + 1.05i
0.170V2[90] = -19 + 1.05i
0.170V2{90] = 19.03[176.8o]
V2 = 112m/s[86.8o] = Velocity of the puck.
To find the speed of the hockey puck before the collision, we need to use the principles of conservation of momentum and conservation of kinetic energy.
Let's use subscripts 1 and 2 to represent the initial and final states, respectively, of the clay and the hockey puck.
1. Conservation of momentum:
Before the collision:
Momentum of clay (initial) = Momentum of hockey puck (initial)
M_clay * V_clay + M_puck * V_puck = (M_clay + M_puck) * V_final
where:
M_clay = mass of clay = 8.34 kg
V_clay = initial velocity of clay = 2.40 m/s
M_puck = mass of hockey puck = 0.170 kg
V_puck = initial velocity of hockey puck (to be determined)
V_final = final velocity of the combined clay and hockey puck = 2.59 m/s
Now we can solve for V_puck:
(8.34 kg * 2.40 m/s) + (0.170 kg * V_puck) = (8.34 kg + 0.170 kg) * 2.59 m/s
19.936 + 0.170V_puck = 21.69226
0.170V_puck = 1.75626
V_puck = 10.324 m/s
Therefore, the speed of the hockey puck before the collision was 10.324 m/s.