Question

Q kardo pls!!!!!!!! ​

Answer 1

Answer:

★ MOMENTUM CONSERVATION

★ ELASTIC COLLISION

Explanation:

★ GIVEN ;

» 2 billiard ball

» Mass of ball ( m ) = 0.05 Kg

» Velocity ( v ) = 6 m / s

★ Impulse ( I ) = ??

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★ IMPULSE ( I ) = Change in Momentum ( ∆ P ) / Time taken ( t )

[ As both balls moving in opposite directions ;

A / c to relative Velocity ;

BOTH the velocities get added ]

★ ∆ P = m1v1 - ( - m2v2 )

[ M1 = M2 ; V1 = V2 ]

{ ∆ P = 2 × MV }

» IMPULSE ( I ) = ∆ P / ∆ t

= 2 × MV / 1 second

= 2 × 0.05 × 6

= 0.6 Kg m / s .

BUT ;

DUE TO IMPULSE ; Both balls rebound on there path.

then

IMPULSE ( I ) = - 0.60

OPTION 3

Answer 2

Given:-Initial Speed of one ball=6m/s

mass of one ball=.05kg

Initial,speed of another ball=6m/s

mass of another ball=0.5kg

To find:-impulse imaprted in balls,under the given condition

Concept:-After collision the speed of the ball will be same,but velocity will he different due to change in direction of motion...

Solution:-

We know, impulse on any body=Force ×Time

also, Force=m(v-u)/T

putting this value of force ,we get

impulse=m(v-u)/T ×T=(mv-mu)

So,we can say impulse is the change in momentum

Before collision, momentum of one ball=0.05×6=0.3

after collision,the ball.will start to move in opposite direction,

so,it's velocity will =-6 (-sign to indicate opposite speed)

momentum in this case=.05×(-6)=-0.3

impulss=change in momentum=mv-mu=-0.3-0.3=-0.6

Hence the momentum imparted on ball will be -0.6

Note:-same case for the second ball