A ball of mass 4 kg moving on a smooth horizontal surface makes an ela...
Given:
- Mass of the first ball (m1) = 4 kg
- Mass of the second ball (m2) = m (unknown)
- Velocity of the first ball before collision (v1) = v
- Velocity of the second ball before collision (v2) = 0 (at rest)
- Velocity of the first ball after collision (v1') = v/4
To find:
The mass of the second ball (m2)
Assumptions:
- The collision is elastic, which means both kinetic energy and momentum are conserved.
Solution:
1. Conservation of Momentum:
- Before the collision, the total momentum is given by:
initial momentum = m1 * v1 + m2 * v2
Since v2 = 0, the equation becomes:
initial momentum = m1 * v1 + 0
- After the collision, the total momentum is given by:
final momentum = m1 * v1' + m2 * v2'
Since v1' = v/4, the equation becomes:
final momentum = m1 * (v/4) + m2 * v2'
- According to the conservation of momentum, initial momentum = final momentum:
m1 * v1 = m1 * (v/4) + m2 * v2'
- Substituting the given values:
4 * v = 4 * (v/4) + m * 0
- Simplifying the equation:
4 * v = v + 0
3 * v = 0
v = 0
2. Conclusion from Momentum Conservation:
- We obtained v = 0, which means the first ball was initially at rest.
- Since the first ball was at rest, it cannot collide with the second ball.
- Therefore, the given scenario is not possible, and we cannot determine the mass of the second ball.
Summary:
- In the given scenario, the mass of the second ball cannot be determined because the initial velocity of the first ball is zero, which implies that there is no collision with the second ball.
A ball of mass 4 kg moving on a smooth horizontal surface makes an ela...
Using conservation of momentum,
initial momentum=final momentum ( in elastic collision)
M1=4 kg, M2=m(unknown), u1=u, u2=0
4u+m(0)=4 (u/4) + mv
v=3u/m is eq.1
In elastic collision, kinetic energy is also conserved.
m1u1^2 + m2u2^2 = m1v1^2 + m2v2^2
4u^2 + m(0)^2 = 4(u/4)^2 + mv^2
on substituting the value of v from eq.1
we get m= 2.4 kg
Hope it helps!!
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