Question

Two particles of masses , and m, in projectile
motion have velocities ví <, respectively at time to
They collide at time to. Their velocities become y, and
at time 2 to, while still moving in air. The value of
1m, + m, 3)-(m, 7 + mū) lis
(a) (m + M₂ )g to (6) 2 (my + M₂) 8 to
(c) / ( + M) 8 to (d) zero
LIIT 01​

Answer 1

Answer:

Explanation:

As there is no external force in the horizontal direction their momentum is changed in the vertical direction only by the gravitation force in time 0 to 2t0

As change in momentum = external force xx time interval

∴(m1→υ1+m2→υ2)-(m1→υ1+m2→υ2)

=(m1+m2)gx(2t0-0)

=2(m1+m2)gt0 .

Answer 2

Given that,

Two particles of masses m₁ and m₂ in projectile motion have velocities v₁ and v₂ respectively at time t = 0. They collide at time t₀. Their velocities become v'₁ and  v'₂ at time  2t₀, while still moving in air.

Find the value of $((m_{1}v_{1}'+m_{2}v_{2}')-(m_{2}v_{1}+m_{2}v_{1})$

Mass of first particle = m₁

Mass of second particle =m₂

Initial velocity of first particle = v₁

Initial velocity of second particle = v₂

Final velocity of first particle = v'₁

Final velocity of second particle = v'₂

Final time = 2t₀

Two particle in projectile motion at t = 0.

They collide at t = t₀.

We know that,

Momentum :

The momentum of the particle is equal to the product of mass and velocity of the particle.

$P=mv$

We need to calculate the value of $((m_{1}v_{1}'+m_{2}v_{2}')-(m_{2}v_{1}+m_{2}v_{1})$

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Put the value into the formula

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

$((m_{1}v_{1}'+m_{2}v_{2}')-(m_{2}v_{1}+m_{2}v_{1})$

This is the change in momentum.

We need to calculate the change in time

Using formula for change in time

$\Delta t=t_{f}-t_{i}$

Put the value into the formula

$\Delta t=2t_{0}-0$

We need to calculate the value of change in momentum

Using formula of change in momentum

$P_{f}-P_{i}=F(t_{f}-t_{i})$

$\Delta  P=F\Delta t$

Here, F = external force

In given condition,

The external force is gravitational force.

So, The change in momentum is

$\Delta P=(m_{1}+m_{2})g\times 2t_{0}$

$\Delta P=2(m_{1}+m_{2})gt_{0}$

Hence, The value of $((m_{1}v_{1}'+m_{2}v_{2}')-(m_{2}v_{1}+m_{2}v_{1})$ is $2(m_{1}+m_{2})gt_{0}$

(b) is correct option.