Momentum, conservation of momentum

In summary: Kinetic energy of block m1 is now 0 so the potential energy is also 0. It is at the bottom of the slope.
  • #1
Jessicaelleig
12
0
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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  • #2
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?
 
  • #3
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.
 

Related to Momentum, conservation of momentum

What is momentum?

Momentum is a physical quantity that describes the amount of motion an object has.

What is conservation of momentum?

Conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant over time, regardless of any external forces acting on the system.

What factors affect momentum?

The two main factors that affect momentum are an object's mass and velocity. The greater an object's mass and velocity, the greater its momentum will be.

How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. The formula for momentum is p = m * v, where p represents momentum, m represents mass, and v represents velocity.

Why is conservation of momentum important?

Conservation of momentum is important because it helps us understand and predict the behavior of objects in motion. It is also a fundamental principle that is used in many areas of physics and engineering, such as in collisions and rocket propulsion.

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