Momentum Problem

[HollowSuperet36]

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Okay so I have the concepts and formulas down. But I still get tripped up on questions that I feel should be very easy. Any help on this one...

A block of mass M starts from rest and slides down a frictionless semi-circular track from a height H as shown below (image shows a perfect circle as the track and the two objects). When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass M. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is:

Highlight for answer.

A. H/4
B. H/2
C. H
D. Independent of H.

Answer is A.

How should I be thinking about this to see this intuitively?

 

[MrNeuro]

Okay so I have the concepts and formulas down. But I still get tripped up on questions that I feel should be very easy. Any help on this one...

A block of mass M starts from rest and slides down a frictionless semi-circular track from a height H as shown below (image shows a perfect circle as the track and the two objects). When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass M. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is:

Highlight for answer.

A. H/4
B. H/2
C. H
D. Independent of H.

Answer is A.

How should I be thinking about this to see this intuitively?

its potential energy is converted into kinetic energy and then undergoes an inelastic collision w/ the putty that has an equal mass (b/c of this its escape velocity is half of the velocity of the object before collision) after this we know that v^2 is proportional to the height hence its H/4 (to prove the proportional thing plug in sqrt2gh into v^2 and solve for the h in potential energy and prove to yourself that the 2 in 2gh cancels out w/ the 1/2 from 1/2mv^2)

math way (which i followed initiall)

mgH = .5 mv^2

v=sqrt 2gh
M*sqrt2gh = 2M V?
V? = 1/2 sqrt2gh

1/2m (V?)^2 = mg Hnew
Hnew= .5 (V?)^2/g
Hnew = 1/2 (1/4*2gH) /g
Hnew= H/4

 

[Ineedhopenow]

I think of it in terms of conservation of energy.

Momentum is conserved, so if mass doubles, velocity has to go down by a factor of 2 (1/2).

KE is proportional to velocity squared, so that means (1/2)^2 = 1/4. That means that KE went down by a factor of 4. Since KE=PE (conservation of energy), PE also goes down by a factor of 4. Therefore height goes down by a factor of 4.

 

[HollowSuperet36]

Excellent, makes much sense. thank you. I forgot to input the value of V when solving in the second half of the problem.

 

[HollowSuperet36]

The other problem that didn't click.

Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed V towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of V/3 in the same direction. What type of collision has occured?

A, Inelastic
B. Elastic
C. Completely Inelastic
D. Cannot be determined from the information given.

B. Elastic

 

[Ineedhopenow]

I initially chose inelastic collision because both balls are traveling the same direction.

However, I think the answer is elastic collision because the velocity is V/3. If it was inelastic, V would be 2/3V. This means that ball 1 and ball 2 are separated, therefore it's an elastic collision.