Collision Question Homework: Masses m1 & m2 in Frictionless Bowl

  • Thread starter Thread starter DMRika
  • Start date Start date
  • Tags Tags
    Collision
Click For Summary
SUMMARY

The discussion focuses on a physics homework problem involving two unequal masses, m1 and m2, in a frictionless bowl. The problem requires algebraic expressions for kinetic energy lost during a collision and the maximum height reached after sliding together. Key points include that if the masses stick together, the kinetic energy lost can be calculated using conservation of momentum and energy principles. For elastic collisions, the total kinetic energy before and after the collision remains constant, and the relationship between the masses can be determined based on their velocities post-collision.

PREREQUISITES
  • Understanding of conservation of momentum
  • Knowledge of kinetic energy calculations
  • Familiarity with elastic and inelastic collisions
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the principles of conservation of momentum in collisions
  • Learn how to calculate kinetic energy in various scenarios
  • Explore the differences between elastic and inelastic collisions
  • Practice solving physics problems involving multiple masses and energy conservation
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of collision problems in a frictionless environment.

DMRika
Messages
2
Reaction score
0

Homework Statement


Two unequal masses m1 and m2 are located in a large, frictionless bowl, and both start at rest. Mass m1 is released at height H and slides to the bottom, where it collides with mass m2. For parts (a) and (b), your answers will be algebraic expressions.

Express your answers only in terms of variables m1, m2, H, g, and mathematical constants, and only in simplest algebraic form.

a. If the two masses stick together in the collision at the bottom of the bowl, find the amount of kinetic energy lost in the collision.

b. If they then slide together up the right-hand side of the bowl, find the maximum height that the two masses reach before sliding back down.

Instead of sticking together, suppose that m1 and m2 collide at the bottom of the bowl perfectly elastically. This applies both to part (c) and (d).

c. Which one of the following statements is now true for the elastic collision?
A. The total K of all masses immediately before collision is less than the total K of all masses immediately after collision.

B.The total K of all masses immediately before collision is the same than the total K of all masses immediately after collision.

C. The total K of all masses immediately before collision is greater than the total K of all masses immediately after collision.

D. Net ∆K of all masses in this perfectly elastic case is the same as in the stuck-together case (part (a)).d. Immediately after the colision, suppose the velocity of m1 rebounds to the left. Which one of the following must be true about the relationship between the two masses?
A. m1 < 1/2m2
B. m1 < m2
C. m1 < 2m2
D. m1 > 1/2m2
E. m1 > m2
F. m1 >2m2

Homework Equations


Not sure

The Attempt at a Solution


I thought it was Vf = m1v1i + m2v2i/m1 +m2 because it is stated that the kinetic energy lost equals the total kinetic energy before the collision (part a).

Didn't attempt parts b, c, and d.
 
Physics news on Phys.org
Hi DMRika! :smile:
DMRika said:
Two unequal masses m1 and m2 are located in a large, frictionless bowl, and both start at rest. Mass m1 is released at height H and slides to the bottom, where it collides with mass m2. For parts (a) and (b), your answers will be algebraic expressions.

Express your answers only in terms of variables m1, m2, H, g, and mathematical constants, and only in simplest algebraic form.

a. If the two masses stick together in the collision at the bottom of the bowl, find the amount of kinetic energy lost in the collision.

I thought it was Vf = m1v1i + m2v2i/m1 +m2 because it is stated that the kinetic energy lost equals the total kinetic energy before the collision (part a).

i] the question asks for the https://www.physicsforums.com/library.php?do=view_item&itemid=132" lost, not for V

ii] you are only allowed to use m1, m2, H, and g in your answer, not v1i and v2i

iii] it does not state that the kinetic energy lost equals the total kinetic energy before the collision (and that isn't true) … do you know what kinetic energy is?

start again :smile:
 
Last edited by a moderator:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
16
Views
5K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K