Question

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 3.05 m/s . A) Calculate the magnitude of the velocity of the puck after a force of 25.5 N directed to the right has been applied for 6.0×10⁻² s B) What is the direction of the velocity of the puck after a force of 25.5 N directed to the right has been applied for 6.0×10⁻² s : (to the left or to the right ) C) If instead, a force of 11.8 N directed to the left is applied from t=0 to t= 6.0×10⁻² s , what is the magnitude of the final velocity of the puck? D) What is the direction of the final velocity of the puck in this case? (to the left, or to the right)

Answer 1

A)
m = mass of the puck = 0.160 kg
u = initial velocity = 3.05 m/s
F = applied force = 25.5 N
t = time = 0.06 s

v = final velocity = ?

(final momentum) = (initial momentum) + (impuls)
m×v = m×u + F×t

Solve for v
v = (m×u + F×t) / m
v = u + F×t/m

v = (3.05 m/s) + (25.5 N)×(0.06 s)/(0.160 kg)
v = 12.61 m/s

B) value is positive, so to the right)

C)
m = mass of the puck = 0.160 kg
u = initial velocity = 3.05 m/s
F = applied force = -11.8 N (note the negative sign)
t = time = 0.06 s

v = final velocity = ?

v = (3.05 m/s) + (-11.8 N)×(0.06 s)/(0.160 kg)
v = -1.375 m/s

D)the final velocity is negative, so to the left