While sliding horizontal on a frictionless surface at a speed of 30 m/s, a hockey puck of mass 0.16 kg is struck by a hockey stick. After leaving the stick, the puck has a speed of 22 m/s in the opposite direction. The magnitude of the impulse given to the puck is _____ N s.
impulse=changemomentum=mass(30+22)
= .16*52 Ns
wait where did you get 52 from?
To find the magnitude of the impulse given to the puck, we can use the impulse-momentum principle, which states that the impulse acting on an object is equal to the change in momentum of the object.
The impulse (J) is given by the equation:
J = Δp = pf - pi
Where J is the impulse, Δp is the change in momentum, pf is the final momentum of the object, and pi is the initial momentum of the object.
The momentum (p) of an object is calculated using the equation:
p = m * v
Where p is the momentum, m is the mass of the object, and v is the velocity of the object.
Given:
Mass of the puck (m) = 0.16 kg
Initial velocity of the puck (vi) = 30 m/s
Final velocity of the puck (vf) = -22 m/s (since it is in the opposite direction)
First, let's calculate the initial momentum (pi) of the puck:
pi = m * vi = 0.16 kg * 30 m/s = 4.8 kg·m/s
Next, let's calculate the final momentum (pf) of the puck:
pf = m * vf = 0.16 kg * (-22 m/s) = -3.52 kg·m/s
Now, let's calculate the change in momentum (Δp):
Δp = pf - pi = (-3.52 kg·m/s) - (4.8 kg·m/s) = -8.32 kg·m/s
The magnitude of the impulse given to the puck is the absolute value of the change in momentum, so:
Impulse (J) = |Δp| = |-8.32 kg·m/s| = 8.32 kg·m/s
Therefore, the magnitude of the impulse given to the puck is 8.32 N·s.