# Use the equations below:

P=mv
Ek=1/2mv^2
2) A linear air track can be used to investigate collisions. Two trolleys or “gliders” are supported on a cushion of air. A glider of mass 0.30kg is stationary in the middle of the track. A second glider of mass 0.25kg and velocity of 0.20ms-1 collides with the first glider and they stick together.
You may assume that this collision is perfectly elastic.
A) What is meant by a “perfectly elastic” collision?
B) Calculate the velocity of the glider combination immediately after the collision.

6 marks

## is it 0.5 * 0.2= 0.11

then 0.11=1/0.55*0.20^2 though i dont think that makes much sense

## Conservation of kinetic Energy.

0.5*M2*V^2 = 0.5*M1*V^2 + 0.5*M2*V^2.
0.5*0.25*0.2^2 = 0.5*0.3V^2+0.5*0.25V^2
0.005 = 0.15V^2 + 0.125V^2,
=.005 = 0.275V^2,
V^2 = 0.0182,
V = 0.135 m/s.

## A) In physics, a perfectly elastic collision refers to a type of collision in which both kinetic energy and momentum are conserved. In other words, when two objects collide elastically, they rebound from each other with no loss of kinetic energy. This implies that the total kinetic energy before and after the collision remains the same, and the total momentum is conserved.

B) To calculate the velocity of the glider combination immediately after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

First, let's calculate the initial momentum of the system before the collision. The initial momentum of the first glider, which is stationary, can be given as:

p1_initial = m1 * v1_initial
= 0.3 kg * 0 m/s
= 0 kg.m/s

The initial momentum of the second glider can be given as:

p2_initial = m2 * v2_initial
= 0.25 kg * 0.20 m/s
= 0.05 kg.m/s

The total initial momentum of the system is the sum of the initial momenta of the two gliders:

total_initial_momentum = p1_initial + p2_initial
= 0 kg.m/s + 0.05 kg.m/s
= 0.05 kg.m/s

Next, since it is a perfectly elastic collision, the total final momentum of the system is equal to the total initial momentum:

total_final_momentum = total_initial_momentum
= 0.05 kg.m/s

Now, let's calculate the total final kinetic energy of the system. Since kinetic energy is given by E_k = 1/2 * m * v^2, we can use the following equation:

total_final_kinetic_energy = 1/2 * m1_final * v1_final^2 + 1/2 * m2_final * v2_final^2

Since the two gliders stick together after the collision, they will have the same final velocity. Let's call this velocity v_final.

m1_final = m1 + m2
= 0.3 kg + 0.25 kg
= 0.55 kg

m2_final = m1_final

Now, substituting everything into the equation, we get:

total_final_kinetic_energy = 1/2 * (m1_final + m1_final) * v_final^2
= m1_final * v_final^2

Since this is an elastic collision, the total final kinetic energy remains the same as the total initial kinetic energy:

total_final_kinetic_energy = total_initial_kinetic_energy

Therefore, we can equate the two equations:

m1_final * v_final^2 = 1/2 * m1 * v1_initial^2 + 1/2 * m2 * v2_initial^2

Now, let's plug in the values:

0.55 kg * v_final^2 = 1/2 * 0.3 kg * 0^2 + 1/2 * 0.25 kg * (0.20 m/s)^2

Simplifying the equation:

0.55 kg * v_final^2 = 0.0025 kg.m^2/s^2

Finally, we can solve for v_final:

v_final^2 = 0.0025 kg.m^2/s^2 / 0.55 kg
v_final^2 = 0.0045 m^2/s^2

Taking the square root of both sides:

v_final = sqrt(0.0045) m/s
v_final ≈ 0.067 m/s

Therefore, the velocity of the glider combination immediately after the collision is approximately 0.067 m/s.