Question

A hockey puck with mass 0.30 kg is sliding along the ice with initial speed of 12.68 m/s. A hockey player is heading toward the puck with his stick in hand. After the player strikes the puck, the puck reverses its direction and is traveling at double its speed before the strike. If the collision occurs in 0.05 s, what is the magnitude of the force the hockey player's stick applied to the puck

Answer 1

F = 228.24 N

Step-by-step explanation:

• According Newton's 2nd Law, the impulse on one object is equal to the change in momentum of that object.
• I = F*Δt = Δp = pf - po (1)

where pf = final momentum = m*vf

p₀ = initial momentum = m*v₀

• Since after the strike, the puck reverses its direction and travels at double its speed before the strike, that means that vf = -2*v₀.
• Replacing in the right side of (1), we have:

`m*v_(f) - m*v_(o) = -2*v_(o) -m*v_(o) = -3*m*v_(o) = -3*0.3kg*12.68m/s = -11.41m/s (2)`

• Replacing Δt = 0.05s, and solving for F in (1):

`F_(net) = (-11.41m/s)/(0.05s) = -228.24 N (3)`

• which means that the force is applied in a direction opposite to the initial velocity of the puck.
• The magnitude of the force is just 228.24 N.