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Elastic/inelastic problems: A 15.0 kg penguin waddling east at a velocity of 7.0 m/s collides with a ...
s.h_math
s.h_math
wrote...
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A month ago
A month ago
Elastic/inelastic problems: A 15.0 kg penguin waddling east at a velocity of 7.0 m/s collides with a ...
Hi there, I was wondering if I could get help with this question. Thanks a lot
A 15.0 kg penguin waddling east at a velocity of 7.0 m/s collides with a stationary 10.0 kg penguin. After the collision the 15.0 kg penguin is traveling at a velocity Of 4.2 m/s 20.0
0
S of E.
a. What is the velocity of the 10.0 kg penguin after collision?
b. is this collision elastic or inelastic?
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duddy
wrote...
#1
Staff Member
A month ago
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.
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- Master of Science in Biology
- Bachelor of Science (Biology)
- Bachelor of Education
s.h_math
wrote...
#2
A month ago
Hey duddy it makes sense but could you write the initial equations to what we are plugging the numbers in? Thankss
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duddy
wrote...
#3
Staff Member
A month ago
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?
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- Master of Science in Biology
- Bachelor of Science (Biology)
- Bachelor of Education
s.h_math
wrote...
#4
A month ago
Yeah for sure thanks a lot!
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Yeah.Yeah.
wrote...
#5
A month ago
you just need number 5?
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s.h_math
wrote...
#6
A month ago
Yeah, as it’s the only number circled.
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