A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck...
A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 3.02 m/s .
part A)
Calculate the magnitude of the velocity of the puck after a force of 25.4 N directed to the right has been applied for 6.0×10−2 s
V= m/s
Homework Answers
Let us consider the right side direction as positive and the left side direction as negative.
Mass of the hockey puck = m = 0.16 kg
Initial velocity of the hockey puck = V1 = 3.02 m/s
Final velocity of the hockey puck = V2
Force on the puck = F = 25.4 N
Time period the force is applied for = T = 6 x 10-2 sec = 0.06 sec
Impulse of the force on the hockey puck = I
I = FT
I = (25.4)(0.06)
I = 1.524 N.s
Impulse is equal to the change in momentum of the object.
I = mV2 - mV1
1.524 = (0.16)V2 - (0.16)(3.02)
V2 = 12.5 m/s
A) Magnitude of velocity of the hockey puck after the force has been applied = 12.5 m/s
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A 0.160-kg hockey puck is moving on an icy, frictionless,
horizontal surface. At t=0 the puck...
A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck...
A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 3.02 m/s . Part A) Calculate the magnitude of the velocity of the puck after a force of 25.4 N directed to the right has been applied for 6.0×10−2 s . V= m/s
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