Momentum, conservation of momentum

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The discussion centers on a physics problem involving a 0.450 kg block colliding elastically with a 1.00 kg mass on a table after descending a frictionless track. Key points include the clarification that momentum conservation does not apply due to external forces like gravity, making it primarily a conservation of energy problem. Participants emphasize calculating potential and kinetic energy at various points to determine speeds after the collision and the heights the blocks reach afterward. The discussion suggests using conservation laws and textbook formulas for elastic collisions to solve for the required speeds and distances. Understanding these principles is crucial for accurately solving the problem.
Jessicaelleig
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Conservation of momentum! help please!?

A m1 = 0.450 kg block is released from rest at the top of a frictionless track h1 = 2.85 m above the top of a table. It then collides elastically with a 1.00 kg mass that is initially at rest on the table, as shown in Figure P6.47.

Figure P6.47.

(a) Determine the speeds of the two masses just after the collision.
speed of m1
m/s
speed of m2
m/s

(b) How high up the track does the 0.450 kg mass travel back after the collision?
m
(c) How far away from the bottom of the table does the 1.00 kg mass land, given that the table is 2.00 m high?
m
(d) How far away from the bottom of the table does the 0.450 kg mass eventually land?
m

Can anyone please give pointers or equations. I am so lost
 
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Jessicaelleig said:
Conservation of momentum! help please!?

A m1 = 0.450 kg block is released from rest at the top of a frictionless track h1 = 2.85 m above the top of a table. It then collides elastically with a 1.00 kg mass that is initially at rest on the table, as shown in Figure P6.47.

Figure P6.47.

(a) Determine the speeds of the two masses just after the collision.
speed of m1
m/s
speed of m2
m/s

(b) How high up the track does the 0.450 kg mass travel back after the collision?
m
(c) How far away from the bottom of the table does the 1.00 kg mass land, given that the table is 2.00 m high?
m
(d) How far away from the bottom of the table does the 0.450 kg mass eventually land?
m

Can anyone please give pointers or equations. I am so lost
First, this is NOT a "conservation of momentum" question because there is an external force (gravity). Momentum of a system is conserved only when there are no external forces. Since you are told that the collision is elastic, it IS a "conservation of energy" problem. You can take potential energy to be 0 at any point so you might as well take it to be 0 at the bottom of the slope.

What is the potential energy of m1= .45 kg block at the top of the track 2.0 m above? What is the kinetic energy of the unmoving block m1? What is its total energy (and so total energy of the system) at that time?

What is the potential energy of block m1 at the bottom of the slope? What is its kinetic energy there?. Momentum is conserved, as well as energy, during the collision. Use both to find the speeds of both blocks immediately after the collision. What is the kinetic energy and total energy of block m1 now?

When the kinetic energy of block m1 is 0, what it its potential energy? So how far up the slope is it?
 
First you'll have to find the speeds of the blocks before collision. (The second one is at rest, this will be the easy one) .
Then apply conservation laws for collision or just use the formulas in the textbook for elastic collision.
 
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