Elastic/inelastic problems: A 15.0 kg penguin waddling east at a velocity of 7.0 m/s collides with a ...

Part A:

Using momentum conservation we have:

Direction \(x\):

(15) (7.0) + 0 = (15) (4.2) cos (-20 °) + (10) v₂ (f) cosθ

(10) v₂ (f) cosθ = (15) (7.0) - (15) (4.2) cos (-20 °) ... (Eq 1)

Direction \(y\):

0 + 0 = (15) (4.2) sin (-20 °) + (10) v₂ (f) sinθ

(10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °) ... (Eq 2)

We divide both equations (1) and (2):

Left side:

(10) v₂ (f) sinθ / (10) v₂ (f) cosθ = tanθ

Right side:

- (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °)

Clearing the angle we have:

θ = tan⁻¹ [ - (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °) ]

θ = 25 °

We use any of the equations to find the speed.

From (2) we have:

(10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °)

v₂ = - (15) (4.2) sin (-20 °) / (10) sin25 °

Answer:

v₂ = 5.1m / s

Note: θ = 25 ° (direction is in the first quadrant, north of east)

Part B:

Initial kinetic energy: K (i) = 0.5 (15) (7.0) ² = 370J (rounded)

Final kinetic energy: KE (f) = 0.5 (15kg) (4.2m / s) ² + 0.5 (10kg) (5.1m / s) ² = 260J (rounded)

Answer:

A difference of 110J, so it was an inelastic collision.

Hey duddy it makes sense but could you write the initial equations to what we are plugging the numbers in? Thankss

The conservation of momentum formula goes like this:

total momentum before = total momentum after

linear momentum = mass * velocity

In other words: \(\rho =m\cdot v\)

I'll break down the first one for you, and hopefully you can apply it to the rest.

Recall:

Direction \(x\):

m*v + m*v = m*v + m*v
1..........2.........3....... ...4

in term 1: (15) (7.0)
in term 2: (10) (0) = 0
in term 3: (15) (4.2) cos (-20°)
in term 4: (10) v₂ (f) cosθ

how's that?

Yeah for sure thanks a lot!

you just need number 5?

Yeah, as it’s the only number circled.