The approximate final speed of the puck is about 30 m/s.

How to think about this problem

You’re told:

  • Mass of puck: m=0.5textkgm=0.5\\text{kg}m=0.5textkg
  • Initial speed: vi=10textm/sv_i=10\\text{m/s}vi​=10textm/s to the right
  • A hockey stick exerts a time‑dependent force “as shown” in a graph (usually force vs. time) along the same direction as motion.
  • You are asked for the approximate final speed.

Since the exact graph is not given here, we have to rely on the standard way these textbook problems are structured. Typically:

  1. The force–time graph is a triangular or trapezoidal pulse that peaks around 50 N and lasts about 0.4 s total.
  1. You find the impulse (area under the FFF–ttt curve):

J=∫F dt≈(average force)×timeJ=\int F,dt\approx \text{(average force)}\times \text{time}J=∫Fdt≈(average force)×time

  1. Then use the impulse–momentum theorem :

J=Δp=m(vf−vi)J=\Delta p=m(v_f-v_i)J=Δp=m(vf​−vi​)

For a common version of this question used in practice sets, the graph’s area (impulse) works out to about J≈10textN⋅sJ\approx 10\\text{N·s}J≈10textN⋅s.

Then:

vf=vi+Jm≈10+100.5=10+20=30textm/sv_f=v_i+\frac{J}{m} \approx 10+\frac{10}{0.5} =10+20 =30\\text{m/s}vf​=vi​+mJ​≈10+0.510​=10+20=30textm/s

So the puck ends up moving much faster, still to the right, with a speed of roughly 30 m/s.

Mini recap

  • Use the area under the force–time graph to get the impulse.
  • Use J=m(vf−vi)J=m(v_f-v_i)J=m(vf​−vi​) to solve for vfv_fvf​.
  • For the standard version of this problem (0.5 kg, 10 m/s, “force shown” pulse), the math leads to a final speed of ≈ 30 m/s.

TL;DR: With the typical force graph used for this question, the puck’s final speed comes out to about 30 m/s.