A ball of mass m, moving with a speed 2v0, collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than 90°.
A ball of mass m, moving with a speed 2v0, collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than 90°.
This is a short answer type question as classified in NCERT Exemplar
Let v1 and v2 are velocities of two balls after collision
According to conservation of momentum
2mvo = mv1+ mv2
2vo= v1+v2
and e= v2-v1/2vo
v2=v1+2voe
2v1=2vo-2evo
V1=vo (1-e) since e<1 so ball will move after collision.
b)by principle of conservation of linear momentum

P=P1+P2
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Using Newton’s formula,
This is a multiple choice answer as classified in NCERT Exemplar
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’/
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 =-
This is a multiple choice answer as classified in NCERT Exemplar
(c) m =150g =3/20kg
Time of contact =0.001s
U=126km/h=
V= -35m/s
Change in momentum of the ball = m (v-u)=
=21/2
F= dp/dt=- = - 1.05
Here – negative sign indicates that force will be opposite to the direction of movement of the ball be
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Physics NCERT Exemplar Solutions Class 11th Chapter Six 2025
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