Question

A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Why does the tackle constitute a perfectly inelastic collision? (b) Calculate the velocity of the players immediately after the tackle and (c) determine the mechanical energy that is lost as a result of the collision. (d) Where did the lost energy go?

Answer 1

Answer:

a) Please see below as the answer is self-explanatory.

b) 2.88 m/s

c) 785. 8 J

d) It is expended like thermal energy, due to internal friction.

Explanation:

a) In a tackle, both players keep emmeshed each other, so it is a perfectly inelastic collision; Immediately after the tackle, both masses behave like they were only one.

b) Assuming no external forces act during the collision, total momentum must be conserved.

As momentum is a vector, the conservation principle must be met by all vector components at the same time.

In our case, as the players move in directions mutually perpendicular, we can decompose the momentum vector along both directions, taking into account that after the collision, the momentum vector will have components along both directions.

So, if we call the W-E axis our X-axis (being the direction towards east as the positive one) , and to the S-N axis our Y -axis (being the northward direction the positive one), we can write the following equations:

pₓ₀ = pₓf ⇒ m₁*v₁ = (m₁+m₂)*vf*cosθ

py₀ = pyf ⇒ m₂*v₂ = (m₁+m₂)*vf*sin θ

where θ, is the angle that both players take regarding the x-axis after the collision (north of east).

Replacing by the values, we have the following equations:

vf*cosθ = (90.0 kg*5.00 m/s) / (90.0 kg + 95.0 kg) = 2.43 m/s (1)

vf*sin θ = (95.0 kg* 3.00 m/s) / (90.0 kg + 95.0 kg) = 1.54 m/s (2)

Dividing both sides:

sin θ / cos θ = tan θ = 1.54 / 2.43 = 0.634

⇒ arc tan (0.634) = 32.3º

Replacing in (1) we have:

vf = 2.43 m/s / cos 32.3º = 2.43 m/s / 0.845 = 2.88 m/s

c) As the collision happens in one dimension, all mechanical energy, before and after the collision, is just the kinetic energy of the players.

Before the collision:

K₀ = 1/2*m₁*v₁₀² + 1/2 m₂*v₂₀²

= 1/2*( ( 90.0) kg*(5.0)²(m/s)² + (95.0)kg*(3.0)(m/s)²) = 1,553 J

After the collision:

Kf = 1/2 *(m₁+ 767.2 Jm₂)*vf² = 1/2*185 kg*(2.88)²(m/s)²= 767.2 J

The mechanical energy lost during the collision is just the difference between the final and initial kinetic energy:

ΔK = Kf - K₀ = 767.2 - 1,553 J = -785.8 J

So, the magnitude of the energy lost during the collision is 785.8 J.

d) This energy is lost during the collision as thermal energy, due to the internal friction between both players.