MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Impulsive Forces in Collision

Two particles...

Question

Two particles of masses $m_{1} a n d m_{2}$ in projectile motion have velocities ${\to v}_{1} a n d {\to v}_{2}$ respectively at time t = 0 and that is the moment when they collide. Their velocities become ${\to v}_{1}^{'} a n d {\to v}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of magnitude of change in total momentum would be

A

$Z e r o$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

$(m_{1} + m_{2}) g t_{0}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

$2 (m_{1} + m_{2}) g t_{0}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

$\frac{1}{2} (m_{1} + m_{2}) g t_{0}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$2 (m_{1} + m_{2}) g t_{0}$

The momentum of the two-particle system, at t = 0 is

${\to p}_{i} = m_{1} {\to v}_{1} + m_{2} {\to v}_{2}$

Collision between the two does not affect the total momentum of the system.

A constant external force $(m_{1} + m_{2}) g$ acts on the system.

The impulse given by this force, in time t = 0 to $t = 2 t_{0} i s (m_{1} + m_{2}) g \times 2 t_{0}$

$∴$ Change in momentum in this interval $= | m_{1} {\to v}_{1} + m_{2} {\to v}_{2} - (m_{1} {\to v}_{1} + m_{2} {\to v}_{2}) | = 2 (m_{1} + m_{2}) g t_{0}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Two particles of masses

m_{1}

m_{2}

in projectile motion have velocities

v_{1}

v_{2}

respectively at time

t = 0

. They collide at time

t_{0}

. Their velocities become

{v^{'}}_{1}

{v^{'}}_{2}

2 t_{0}

while still moving in air. The value of

[(m_{1} {v^{'}}_{1} + m_{2} {v^{'}}_{2}) - (m_{1} v_{1} + m_{2} v_{2})]

Q. A nucleus moving with velocity

\to v

α - p a r t i c l e

. Let the velocities of the

α - p a r t i c l e

and the remaining nucleus be

{\to v}_{1}

{\to v}_{2}

and their masses be

m_{1}

m_{2}

Q. Two particles

1

2

are moving with constant velocities

{\to v}_{1} = 4^i - 3^j m/s

{\to v}_{2} = b^i -^j m/s

respectively. The position vectors of the particles at time

t = 0

{\to r}_{1} = 5^i + 2^j m

{\to r}_{2} = - 4^i - 4^j m

. If they collide, which of the following is correct. (

b

Q. Two particles having position vectors

\to r = (3 \to i + 5 j) m

{\to r}_{2} = (- 5 i + 3 j) m

are moving with velocities

{\to V}_{1} = (4^i - 4^j) m s^{- 1}

{\to V}_{2} = (a^i - 3^j) m s^{- 1}

. If they collide after

2

seconds , the value of

a

Q. STATEMENT-1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.

STATEMENT-2: If the observer and the object are moving at velocities

{\to V}_{1}

{\to V}_{2}

respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is

{\to V}_{2} - {\to V}_{1}

.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Momentum Returns

PHYSICS

Watch in App

explore_more_icon

Explore more

Impulsive Forces in Collision

Standard XI Physics

Join BYJU'S Learning Program

CrossIcon