Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V as shown in fig. If the collision is elastic, which of the following (Fig. 6.4) is a possible result after collision?
This is a multiple choice answer as classified in NCERT Exemplar
(b) When two bodies of equal masses collides elastically, their velocities are interchanged.
When ball 1 collides with ball 2 then velocity of ball1, v1 becomes zero and velocity of ball 2, v2 becomes v.
V1=0, V2=v
When ball 2 collides will ball 3 v2 =0, v3=v
<p><span data-teams="true">This is a multiple choice answer as classified in NCERT Exemplar</span></p><p> (b) When two bodies of equal masses collides elastically, their velocities are interchanged.</p><p>When ball 1 collides with ball 2 then velocity of ball1, v<sub>1</sub> becomes zero and velocity of ball 2, v<sub>2</sub> becomes v.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745403375phpKdc4rB_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745403375phpKdc4rB.jpeg" alt="" width="537" height="210"></picture></div></div><p>V<sub>1</sub>=0, V<sub>2</sub>=v</p><p>When ball 2 collides will ball 3 v<sub>2</sub> =0, v<sub>3</sub>=v</p>
This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”
Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1=………….2
From eqn 1 and 2
So v1 = v’/=v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path
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This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”
Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1=………….2
From eqn 1 and 2
So v1 = v’/=v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path, followed will change and the bullet reaches at point B instead of A
(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.
This is a multiple choice answer as classified in NCERT Exemplar
(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
Hence total work done =-mgL + mgL=0
As the point of application of the contact forces does not move hence work done by reaction forces will be zero.
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