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

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.