Amplified Rebound Height

In summary: So, the only force that is exerted on ball M is the normal force.In summary, we have two small rubber balls of masses m and M dropped from a height h above a hard floor. The heavier ball M reaches the floor first and bounces elastically, while the lighter ball m collides with it at ground level. Using the equations for final velocities after an elastic collision, we can express the height to which the ball of mass m rises after the collision in terms of h, m, and M. We can also use Newton's 3rd law and the equation for velocity to find the position of the ball.
  • #1
JennV
23
0

Homework Statement



Amplified Rebound Height Two small rubber balls are dropped from rest at a height h above a hard floor. When the balls are released, the lighter ball (with mass m) is directly above the heavier ball (with mass M). Assume the heavier ball reaches the floor first and bounces elastically; thus, when the balls collide, the ball of mass M is moving upward with a speed v and the ball of mass m is moving downward with essentially the same speed.

In terms of h, find the height to which the ball of mass m rises after the collision. (Use the equations for final velocities after elastic collision of two objects with masses m_1 and m_2:
v1,f = (m1-m2 / m1+m2)v1 + (2m2 / m1+m2)v2
v2,f = (2m1 / m1+m2)v1 + (m2-m1 / m1+m2)v2
and assume the balls collide at ground level.)

Express your answer in terms of h, m and M.

Homework Equations


The Attempt at a Solution



So something like this?

|
| m (the lighter ball)
|
V
^
|
| M (the heavier ball)
|

AFTER:
The lighter ball bounces back up (elastic) after the collision, so I'm trying to find the height that it bounces back up before it hits the ground?

But how would I find the expression of h? =S
Thanks in advance.
 
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  • #2
First of all, v=sqrt(2*g*h)

Since they collide at ground level, we could assume that the velocity of ball M is zero after the collision. Thus, the velocity,v1, of ball m after the collision is

v1^2=(m+M)V^2/m

Then use Newtons' 3rd law to find the position of the ball. The only force that is exerted on that ball is gravity.
 

1. What is amplified rebound height?

Amplified rebound height is a term used to describe the increased height that an object bounces back to after being released or dropped from a certain height. It is the result of energy being stored and released through the elastic properties of the object and the surface it is bouncing on.

2. How is amplified rebound height measured?

Amplified rebound height can be measured by dropping an object from a known height onto a flat surface and measuring the height it bounces back to. This measurement can be repeated multiple times and an average can be taken to get a more accurate result.

3. What factors affect amplified rebound height?

Several factors can affect amplified rebound height, including the material and shape of the object, the surface it is bouncing on, and the force with which it is dropped or released. Other factors such as air resistance and temperature can also play a role.

4. How does amplified rebound height relate to energy?

Amplified rebound height is directly related to energy. The object gains potential energy as it is lifted to a certain height and this energy is then converted into kinetic energy as it bounces back. The greater the amount of energy that is stored and released, the higher the amplified rebound height will be.

5. What are some real-life examples of amplified rebound height?

Some real-life examples of amplified rebound height include bouncing a ball on a hard surface, jumping on a trampoline, and the rebound of a spring. These all involve the conversion of energy and the resulting increase in height of the object after it bounces back.

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