Force and Laws of Motion

Conservation of Momentum - Numerical Problems

Question 1: Find the recoil velocity of a gun having mass equal to 5 kg, if a bullet of 25gm acquires the velocity of 500m/s after firing from the gun.

Answer: Here given,

Mass of bullet (m1) = 25 gm = 0.025 kg
Velocity of bullet before firing (u1) = 0
Velocity of bullet after firing (v1) = 500 m/s
Mass of gun (m2) = 5 kg
Velocity of gun before firing, (u2) = 0
Velocity of gun after firing = ?

We know that,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`
`=>0.025\ kg xx0+5\ kgxx0` `=0.025\ kg xx 500 m//s + 5\ kg xxv_2`
`=>0=12.5\ kg\ m//s+5\ kg xx v_2`
`=>5\ kg xx v_2 = -12.5\ kg\ m//s`
`=>v_2=(-12.5\ kg\ m//s)/(5\ kg)`
`=>v_2=-2.5\ m//s`

Thus, recoil velocity of gun is equal to 2.5 m/s. Here negative (- ve) sign shows that gun moves in the opposite direction of bullet.


Question 2: A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s, find the velocity of bullet.

Answer: Here we have,

Mass of bullet, m1 = 5 gm = 5/1000 kg = 0.005 kg
Mass of pistol, m2 = 1.5 kg
Recoil velocity of pistol v2 = 1.5 m/s
Velocity of bullet v1 =?

Since, before firing the bullet and pistol are in rest, thus

Initial velocity of bullet, u1 =0

And initial recoil velocity of pistol, u2 =0

We know that,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`
`=>0.005\ kgxx0+1.5\ kg xx0` `=0.005\ kgxx v_1+1.5\ kg xx 1.5m//s`
`=>0=0.005\ kg xx v_1+2.25\ kg\ m//s`
`=>0.005\ kg xx v_1 = -2.25\ kg\ m//s`
`=>v_1=(-2.25\ kg\ m//s)/(0.005\ kg)`
`=>v_1=-450\ m//s`

Thus, velocity of bullet = 450 m/s, here negative sign with velocity of pistol shows that, bullet moves in the opposite direction of pistol.


Question 3: A boy of 50 kg mass is running with a velocity of 2 m/s. He jumps over a stationary cart of 2 kg while running. Find the velocity of cart after jumping of boy.

Answer: Here given,

Mass (m1) of boy = 50 kg
Initial Velocity (u1) of boy = 2 m/s
Mass (m2) of cart = 2 kg
Initial Velocity (u2) of cart = 0
Final velocity of cart (v2) =?

Since, boy jumped over cart thus, thus the final velocity (v1) of boy will be equal to that of the cart.

Therefore, v1 = v2

We know that,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`
`=>50\ kg xx2ms^(-1)+2\ kg xx 0` `=50\ kg xx v_1+2\ kg xx v_2`

∵ `v_1=v_2`
`:. 100\ kg\ m//s=50\ kgxxv_2+2\ kgxxv_2`
`=>100\ kg\ m//s = v_2(50\ kg+2\ kg)`
`=>100\ kg\ m//s=v_2xx52\ kg`
`=>v_2 = (100\ kg\ m//s)/(52\ kg)`
`=> v_2 = 1.92\ m//s`

Therefore, velocity of cart after jumping of boy over it is equal to 1.92 m/s. Since, velocity has positive sign, thus, cart will go in the same direction of boy.


Question 4: While playing football match, Kris collided and got entangled with Tom who was playing for opposite team and running from opposite side. The mass of Kris was 40 kg and the mass of Tom was 60 kg. If Tom was running with a velocity of 3m/s and Kris was running with a velocity of 4 m/s, find the velocity and direction of both of the players after collision assuming other forces were negligible.

Answer:Given,

Mass of Kris (m1) = 40 kg
Initial velocity of Kris (u1) = 4 m/s
Mass of Tom (m2) = 60 kg
Initial velocity of Tom (u2) = 3 m/s
Final velocity and direction of both of the player after collision =?

Let final velocity of both of the players after collision = v

Let Kris was coming from left and Tom was coming from right.

Let the velocity of Kris is positive, therefore velocity of Tom will be negative as both were running in opposite directions.

Thus, initial velocity of Kris (u1) = 4 m/s
And the initial velocity of Tom (u2) = - 3 m/s

We know that,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`
`=>40\ kg xx 4m//s +60\ kg xx (-3m//s)` `=40\ kg xxv +60\ kg xxv`
(Because entangled after collision both the player got same velocity)
`=>40\ kg xx 4m//s +60\ kg xx (-3m//s)` `=40\ kg xx v+ 60\ kg xxv`
`=>160\ kg\ m//s-180\ kg\ m//s ` `=v(40\ kg+60\ kg)`
`=>-20\ kg\ m//s = vxx100\ kg`
`=>v=(-20\ kg\ m//s)/(100\ kg)`
`=> v = -0.2m//s`

Thus, velocity of both the player would become – 0.2 m/s. Negative velocity shows that they would go from right to left after collision.



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