1. a particle A of mass 150g lies at rest on a smooth horizontal surface. a second particle B of mass 100g is projected along the surface with a speed of u m.s and collides directly with A. on collision the masses coalesce and moves on with a sped of 4 m.s . find the value of u?
This is a conservation of linear momentum problem.
Please show your work for further assistance. Someone here will critique your thinking.
To find the value of u, we need to apply the principle of conservation of momentum.
1. The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision, provided no external forces are acting on it.
2. Let's denote the initial velocity of particle B as u (m/s).
3. The initial momentum of particle B can be calculated using the formula:
initial momentum of B = mass of B × velocity of B
initial momentum of B = 100g × u (multiply the mass in grams by the velocity in m/s)
4. Since particle A is at rest, its initial momentum is zero. The total initial momentum of the system is the sum of the initial momentum of particle A and the initial momentum of particle B.
5. After the collision, the two particles coalesce and move together with a speed of 4 m/s. The final momentum of the system can be calculated using the formula:
final momentum = combined mass × final velocity
6. The combined mass of particle A and B is (150g + 100g).
7. According to the principle of conservation of momentum, the initial momentum of the system should be equal to the final momentum of the system.
8. Setting up the equation: Initial momentum = final momentum
(100g × u) + (150g × 0) = (150g + 100g) × 4
Note: We set the initial momentum of particle A as zero, as it is at rest.
9. Simplifying the equation, we have:
100g × u = 250g × 4
10. Now, we can solve for u:
100u = 1000
u = 10 m/s
Therefore, the value of u is 10 m/s.