Science physics
A ball A moving with kinetic energy E makes a head on elastic collision with a stationary ball with mass n times that as A . What is the potential enery stored during collision ?
The momentum does not change but the balls are together during the collision so
initial ke = (1/2) m V1^2 = total energy
initial momentum = m V1
final momentum = (n+1)m V2
so
V2 = V1/(n+1)
so ke during collision
= (1/2) (n+1)m V2^2
= (1/2) (n+1)m V1^2/(n+1)^2
= (1/2)m V1^2/(n+1)
the total energy is constant if elastic so
(1/2)m V1^2 = (1/2)mV1^2/(n+1) + pe
pe = (1/2)mV1^2 [1-1/(n+1]
= (1/2)m V1^2 [n/(n+1)]
= initial Ke [ n/(n+1)]
if n is huge, it all gets stored because the balls stop.
if n is tiny, it goes to zero because the original ball keeps going at just about the same speed.
.... and if you have a physics question type PHYSICS, not Blue bells....,
To determine the potential energy stored during the collision, let's break down the problem step by step.
First, let's define the variables:
- Ball A has a mass of m_A and moves with a kinetic energy of E.
- Ball B has a mass of n * m_A (n times the mass of A) and is stationary.
During an elastic collision, both kinetic energy and momentum are conserved.
Step 1: Conservation of Kinetic Energy
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
1/2 * m_A * v_A^2 (kinetic energy of A before collision) = 1/2 * m_A * v_Af^2 (kinetic energy of A after collision) + 1/2 * (n * m_A) * v_Bf^2 (kinetic energy of B after collision)
Since Ball B is stationary, v_Bf = 0.
1/2 * m_A * v_A^2 = 1/2 * m_A * v_Af^2
Step 2: Conservation of Momentum
Momentum is also conserved in an elastic collision.
m_A * v_A (momentum of A before collision) = m_A * v_Af (momentum of A after collision) + (n * m_A) * v_Bf (momentum of B after collision)
Since Ball B is stationary, v_Bf = 0.
m_A * v_A = m_A * v_Af
Step 3: Solving for Final Velocity
Using the equations from Steps 1 and 2, we can solve for v_Af and v_Bf in terms of v_A:
1/2 * m_A * v_A^2 = 1/2 * m_A * v_Af^2
m_A * v_A = m_A * v_Af
Canceling out the common factor (m_A) in both equations:
v_A^2 = v_Af^2
v_A = v_Af
This means that the final velocity of Ball A is equal to its initial velocity, which makes sense since the collision is elastic.
Step 4: Potential Energy Calculation
Now, to calculate the potential energy stored during the collision, we need to know the definition of potential energy in the context of this problem.
Potential energy is the energy stored due to the position or configuration of an object.
In this case, since we have an elastic head-on collision, there is no change in the configuration or position of the balls. As a result, there is no potential energy stored during the collision.