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Slapshot: impulse, release speed, energy, power

SPH4U Grade 12 Physics · Energy and Momentum

At practice, a 165 g puck sits at rest on the ice. A player's slapshot drives the blade through it with an average force of 132 N over a contact lasting 25.0 ms, and the puck leaves the blade along the ice with friction negligible during the strike.

Given
  • m = 165 g — Mass of the puck
  • F = 132 N — Average force of the blade
  • Δt = 25 ms — Contact time
Determine
  1. (a)the impulse the blade delivers
  2. (b)the speed of the puck as it leaves the blade
  3. (c)the kinetic energy it carries away
  4. (d)the average power the player delivers during the contact
Step 1 of 4(a) · solve for Impulse (change in momentum)

Impulse is force held for a time: J = FΔt. The contact is quoted in milliseconds and the formula wants seconds, so 25.0 ms enters as 0.0250 s — skip that and the impulse is a thousand times too generous.

Rearranged for J
J=F ΔtJ = F \, \Delta t
Your values, in your units
J=(132 N) (25 ms)J = \left( 132\ \text{N} \right) \, \left( 25\ \text{ms} \right)
Converted to base units
J=(132 N) (0.025 s)J = \left( 132\ \text{N} \right) \, \left( 0.025\ \text{s} \right)
Answer
J=3.3 kg⋅m/sJ = 3.3\ \text{kg}{\cdot}\text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Impulse (J = FΔt) solver →

Step 2 of 4(b) · solve for Velocity

The impulse–momentum theorem says J equals the CHANGE in momentum, and because the puck started from rest the change is the whole momentum: p = 3.30 kg·m/s. Divide by the mass in kilograms — 0.165, not 165 — and the release speed is exactly 20.0 m/s, a 72 km/h shot.

Rearranged for v
v=pmv = \tfrac{p}{m}
3.3 kg·m/scarried from step 1
Your values, in your units
v=(3.3 kg⋅m/s)(165 g)v = \tfrac{\left( 3.3\ \text{kg}{\cdot}\text{m/s} \right)}{\left( 165\ \text{g} \right)}
Converted to base units
v=(3.3 kg⋅m/s)(0.165 kg)v = \tfrac{\left( 3.3\ \text{kg}{\cdot}\text{m/s} \right)}{\left( 0.165\ \text{kg} \right)}
Answer
v=20 m/sv = 20\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Linear Momentum (p = mv) solver →

Step 3 of 4(c) · solve for Kinetic energy

Momentum and energy are different books on the same event: p grew linearly with the contact, but Eₖ = ½mv² grows with the SQUARE of the speed the contact bought. Doubling the contact time would double the momentum and quadruple this 33.0 J.

Rearranged for Eₖ
Ek=12mv2E_k = \tfrac{1}{2} m v^{2}
20 m/scarried from step 2
Your values, in your units
Ek=12 (165 g) (20 m/s)2E_k = \tfrac{1}{2} \, \left( 165\ \text{g} \right) \, \left( 20\ \text{m/s} \right)^{2}
Converted to base units
Ek=12 (0.165 kg) (20 m/s)2E_k = \tfrac{1}{2} \, \left( 0.165\ \text{kg} \right) \, \left( 20\ \text{m/s} \right)^{2}
Answer
Ek=33 JE_k = 33\ \text{J}

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 4 of 4(d) · solve for Power

All 33.0 J passed to the puck inside 25.0 ms, so the delivery rate is 1320 W — nearly two horsepower, sustained for a fortieth of a second. Short and violent is what "power" measures; the joules were never large.

Rearranged for P
P=WtP = \frac{W}{t}
33 Jcarried from step 3
Your values, in your units
P=(33 J)(25 ms)P = \frac{\left( 33\ \text{J} \right)}{\left( 25\ \text{ms} \right)}
Converted to base units
P=(33 J)(0.025 s)P = \frac{\left( 33\ \text{J} \right)}{\left( 0.025\ \text{s} \right)}
Answer
P=1.32 kWP = 1.32\ \text{kW}

Carried onward at full precision, not this rounded figure.

Open the Power (P = W/t) solver →

Therefore the blade delivers 3.30 N·s of impulse, sending the 165 g puck away at 20.0 m/s (72 km/h) with 33.0 J of kinetic energy — energy transferred at an average of 1320 W while blade and puck touched.

Why this order

The chain walks the two great bookkeeping systems of mechanics in the order the physics dictates. Impulse comes first because F and Δt are the only things the collision itself offers; momentum follows because J = Δp is a theorem, not a coincidence — it is Newton's second law multiplied by time. Energy comes third and must be COMPUTED from the speed rather than taken from the impulse, because there is no impulse–energy shortcut: the same 3.30 N·s delivered to a heavier puck would produce the same momentum but less speed and therefore less energy. Momentum is what the force-time product buys; where the energy lands depends on what received it. Keeping those ledgers separate is the entire discipline of collision problems, and the reason this chain refuses to jump from step 1 to step 3 directly.

The engineering lives in step 4's ratio of scales. Thirty-three joules is a mug of coffee raised a few metres — trivial energy — but compressed into 25 ms it becomes 1320 W, which is why a slapshot can dent a goalie mask while a slow push with the same energy would not. Every impact tool exploits the same compression: a hammer, a piledriver, an airbag reading the trade in reverse. The airbag is the instructive twin — it cannot change the momentum your body must lose, so it stretches Δt from milliseconds toward a tenth of a second, and F = J/Δt falls by the same factor. Impulse fixed, time negotiable, force the consequence: that one sentence is most of the safety engineering ever built around collisions.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.