Solving Elastic Collisions: 25g & 10g Objects

AI Thread Summary
In the discussion on solving elastic collisions, a 25g object collides with a 10g object, and the goal is to find their velocities post-collision. The initial momentum equation was incorrectly simplified, leading to an erroneous calculation of 2.57 cm/s for one object's velocity. A participant pointed out the mistake in simplifying the momentum equation and suggested isolating one variable to correctly substitute it into the conservation of momentum equation. The correct approach involves properly handling the momentum equations to derive the final velocities. Clarifying the simplification process is crucial for solving the problem accurately.
kimlu
Messages
11
Reaction score
0

Homework Statement


A 25g object moving to the right at 20cm/s overtakes and collides elastically with a 10g object moving in the same direction at 15 cm/s. Find the velocity of each object after the collision


Homework Equations


Ʃp=Ʃp'
v1+v1'=v2+v2'


The Attempt at a Solution


First I tried to find one one of the v primes by using the momentum equation

25(20)+10(15)=25v1'+10v2' and then simplifying it to
500+150=250(v1'+v2')
650-250v1'=v2'

Afterwards I plugged this in for v2' in v1+v1'=v2+v2' and solved through to get 2.57 cm/s for v1' which I know is incorrect. I know I did something wrong in the step above but I don't know how to approach this.
 
Physics news on Phys.org
kimlu said:

Homework Statement


A 25g object moving to the right at 20cm/s overtakes and collides elastically with a 10g object moving in the same direction at 15 cm/s. Find the velocity of each object after the collision


Homework Equations


Ʃp=Ʃp'
v1+v1'=v2+v2'


The Attempt at a Solution


First I tried to find one one of the v primes by using the momentum equation

25(20)+10(15)=25v1'+10v2' and then simplifying it to
500+150=250(v1'+v2')
650-250v1'=v2'

Afterwards I plugged this in for v2' in v1+v1'=v2+v2' and solved through to get 2.57 cm/s for v1' which I know is incorrect. I know I did something wrong in the step above but I don't know how to approach this.

The red part is wrong.

ehild
 
ehild said:
The red part is wrong.

ehild

I'm not quite sure how to simplify that part it would help if you would tell me how :c
 
kimlu said:
I'm not quite sure how to simplify that part it would help if you would tell me how :c

25v1'+10v2' is not 250(v1'+v2'). Isolate v2' from the equation 25v1'+10v2'=650, and plug in for v2' in v1+v1'=v2+v2'.

ehild
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top