Collision Question Homework: Masses m1 & m2 in Frictionless Bowl

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In summary, two unequal masses, m1 and m2, located in a frictionless bowl, start at rest. Mass m1 is released at height H and slides to the bottom where it collides with mass m2. In part (a), if the two masses stick together in the collision, the amount of kinetic energy lost can be expressed as a simple algebraic expression in terms of m1, m2, H, and g.
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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.
 
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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:
 
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Related to Collision Question Homework: Masses m1 & m2 in Frictionless Bowl

1. What is the purpose of the "Collision Question Homework: Masses m1 & m2 in Frictionless Bowl"?

The purpose of this homework question is to test your understanding of collisions between two masses in a frictionless environment, specifically in a bowl-shaped container. It also aims to assess your ability to apply the laws of physics, such as conservation of momentum and energy, to solve problems related to collisions.

2. Can you explain the concept of a frictionless bowl in this homework question?

A frictionless bowl refers to a theoretical scenario where there is no friction or resistance present between the two masses and the surface of the bowl. In this situation, the only forces acting on the masses are their own weight and any external forces applied.

3. How do I determine the velocities of the two masses after they collide in a frictionless bowl?

To determine the velocities of the two masses after a collision in a frictionless bowl, you can use the law of conservation of momentum, which states that the total momentum of a system before and after a collision is constant. You can also use the law of conservation of energy, which states that the total energy of a system before and after a collision is also constant.

4. What are the equations that I need to use in solving this homework question?

The equations that you need to use in solving this homework question include the equations for conservation of momentum (m1v1 + m2v2 = m1v1' + m2v2') and conservation of energy (1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2). You may also need to use the equations for calculating kinetic energy (KE = 1/2mv^2) and potential energy (PE = mgh).

5. What are some common mistakes to avoid when solving this homework question?

Some common mistakes to avoid when solving this homework question include forgetting to account for the direction of velocities, using incorrect values for mass or velocity, and not considering the conservation of energy. It is also important to carefully label and keep track of all the variables and equations used in the solution.

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