Momentum and Collisions with a Spring

In summary, there is a problem involving a frictionless table, a compressed spring, and two blocks with different masses. The velocities of the blocks after the collision are calculated using the equations 1/2kx^2 = 1/2mv^2 and m1v1 + m2v2 = m1v1 + m2v2. The amount of energy lost is calculated using the equation Change in KE = KEfinal-KEinitial = 1/2mv^2 - 1/2mv^2. However, there is confusion about the distance between the impact points on the ground as it is not clear how the 5kg block will hit the ground after the collision. More information is needed to
  • #1
catenn
18
0

Homework Statement


There is a frictionless table 1 meter above the ground. A spring is on one end of the table with a constant of 200N*m that is compressed .5m by a block of 5kg. The block of 5kg is released and then collides with a 2kg block also sitting on the table.
a. Find the velocity of the 5kg block after it is released.
b. Find the velocity of the 2kg block after the colllision.
c. Find the amount of energy lost.
d. Find the distance between the impact points on the ground.

Homework Equations


a. 1/2kx^2 = 1/2mv^2
b. m1v1 + m2v2 = m1v1 + m2v2
c. Change in KE = KEfinal-KEinitial = 1/2mv^2 - 1/2mv^2
d. Projectile Motion? initial velocity off of the table at 7.9m/s for the 2kg and what for the 5kg?

The Attempt at a Solution


a. 3.16m/s
b. 7.9m/s
c. Change in KE = KEfinal-KEinitial = 1/2mv^2 - 1/2mv^2 = 1/2(5)(3.16^2) - 1/2(2)(7.9^2) = 37.446J
d. I was confused at this. I do not see how the 5kg block will hit the ground if it stops after the collision, or does it? If not then it will hit the spring again and be launched with a new velocity after the 2kg block hits?


I need help mostly on part d and think I have a-c although if I'm wrong please tell me. Thanks!
 
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  • #2
The 5kg block will not stop. The problem is faulty, or has not been stated completely. There is no way you can do parts b, c, and d without making an assumption about energy loss in the first place. Was anything said about the nature of the collision? Is it head on or at an angle (assume head on unless otherwise stated)? Was a coefficient of restitution given?
 
  • #3


Great job on parts a-c! Your solutions look correct. For part d, you are correct in thinking that the 5kg block will not hit the ground after the collision. It will hit the spring again and be launched with a new velocity. To find the distance between the impact points on the ground, we can use the equation for projectile motion:

d = (v^2 sin 2θ)/g

where v is the initial velocity off the table (which we can calculate using the conservation of momentum equation in part b), θ is the angle of launch (which we can calculate using trigonometry), and g is the acceleration due to gravity (9.8 m/s^2).

Hope this helps! Keep up the good work in your studies of momentum and collisions.
 

1. What is momentum?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is momentum conserved in a collision with a spring?

In a collision with a spring, momentum is conserved because the total momentum of the system before and after the collision remains the same. This means that the sum of the momentum of the objects involved in the collision remains constant.

3. What is a spring constant?

A spring constant is a measure of the stiffness of a spring and is represented by the letter k. It is a constant value that relates the force exerted by a spring to the distance it is stretched or compressed.

4. How does a spring's spring constant affect momentum in a collision?

The spring constant affects momentum in a collision by determining how much force is exerted by the spring on the objects involved. A higher spring constant means a stronger force and therefore a greater change in momentum in the objects.

5. Can momentum be transferred from a spring to an object?

Yes, momentum can be transferred from a spring to an object through collisions. When an object collides with a spring, the spring exerts a force on the object, causing a change in its momentum. This transfer of momentum is what allows springs to be used in various applications, such as in car shocks and pogo sticks.

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