1. (a) What is an elastic collision ? Derive expres-
sions for final velocities in case of an elastic
collision between two bodies of unequal
masses in one dimension. What happens to the
final velocities if a heavier body collides with
a lighter body at rest?
1 +6+2
(6) Two balls each of mass m moving in opposite
directions with speed y undergo headon colli-
sion. Calculate the velocities of two balls after
the collision
3.5
(c) State the principle of conservation of linear
momentum. Apply it to find an expression
for the instantaneous final velocity of a rocket
due to changing mass during its motion. 1+7.5
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(a)
★An Elastic Collision is a collision in which there is NO NET LOSS IN THE KINETIC ENERGY IN THE SYSTEM as a result of the collision. Both momentum and kinetic energy are conserved quantities in elastic collisions.
Consider two bodies of mass m1 and m2, moving with an initial velocity of u1 and u2, respectively.
Now, apply Kinetic Energy conservation:
KE1 + KE2 = KE1' + KE2'
(m1(u1)²)/2 + (m2(u2)²)/2 = (m1(v1)²)/2 + (m1(v1)²)/2
Multiply both sides by 2, we get:
→ m1(u1)² + m2(u2)² = m1(v1)² + m1(v1)²
this is eqn. (1)
where u1 & u2 are the initial velocities (before collision), and v1 & v2 are the final velocities (after collision), respectively.
Similarly Momentum conservation:
P1 + P2 = P1' + P2'
m1u1 + m2u2 = m1v1 + m2v2
This gives eqn. (2)
On Solving eqn. (1) & (2), we will get v1 & v2 as the final velocities of m1 & m2.
If a heavier body collides with a lighter body at rest, then:
Since the lighter body is initially at rest, then it's initial velocity will be equals to zero, "u = 0".
Apply the same steps, and solve.
Thankyou!!!
★An Elastic Collision is a collision in which there is NO NET LOSS IN THE KINETIC ENERGY IN THE SYSTEM as a result of the collision. Both momentum and kinetic energy are conserved quantities in elastic collisions.
Consider two bodies of mass m1 and m2, moving with an initial velocity of u1 and u2, respectively.
Now, apply Kinetic Energy conservation:
KE1 + KE2 = KE1' + KE2'
(m1(u1)²)/2 + (m2(u2)²)/2 = (m1(v1)²)/2 + (m1(v1)²)/2
Multiply both sides by 2, we get:
→ m1(u1)² + m2(u2)² = m1(v1)² + m1(v1)²
this is eqn. (1)
where u1 & u2 are the initial velocities (before collision), and v1 & v2 are the final velocities (after collision), respectively.
Similarly Momentum conservation:
P1 + P2 = P1' + P2'
m1u1 + m2u2 = m1v1 + m2v2
This gives eqn. (2)
On Solving eqn. (1) & (2), we will get v1 & v2 as the final velocities of m1 & m2.
If a heavier body collides with a lighter body at rest, then:
Since the lighter body is initially at rest, then it's initial velocity will be equals to zero, "u = 0".
Apply the same steps, and solve.
Thankyou!!!
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