Answer:
a. 13.96 m/s in the +x direction
b. -1.76 m/s that is 1.76 m/s in the -x direction
Explanation:
(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.
Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = + 37.0 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck = + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck
So, making v₂ subject of the formula, we have
v₂ = v₁ + Ft/m
substituting the values of the variables into the equation, we have
v₂ = v₁ + Ft/m
v₂ = 2.40 m/s + 37.0 N × 0.050 s/0.160 kg
v₂ = 2.40 m/s + 1.85 Ns/0.160 kg
v₂ = 2.40 m/s + 11.56 m/s
v₂ = 13.96 m/s
(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?
Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = - 13.3 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck = + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck
So, making v₂ subject of the formula, we have
v₂ = v₁ + Ft/m
substituting the values of the variables into the equation, we have
v₂ = v₁ + Ft/m
v₂ = 2.40 m/s + (-13.3 N) × 0.050 s/0.160 kg
v₂ = 2.40 m/s - 0.665 Ns/0.160 kg
v₂ = 2.40 m/s - 4.156 m/s
v₂ = -1.756 m/s
v₂ ≅ -1.76 m/s