A 0.160 kg hockey puck is moving on an icy, frictionless, horizontal surface. At t = 0 the puck is moving to theright at 2.40 m/s. (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.
a. +x direction.
b. +y direction.
c. -x direction.
d. -y direction.
(b) If instead, a force of 13.3 Ndirected to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?
a. +x direction.
b. +y direction.
c. -x direction.
d. -y direction.

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

newton's first law of motion

force tends to be low and long diustance masses are high