A ball of mass 5 kilo gram moving with velocity 3 metre per second strikes A ball of mass 2 kilogram kept at rest. If the lighter ball moves with a velocity 2 metre per second after the collision find the velocity of the heavier ball ?
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We use conservation of kinetic energy as we cannot use conservation of linear momentum as direction of the ball after collision is not known
KE before collision =12=12m.102+0m.102+0
=50mJ=50mJ
KE after collision =12=12m×82+12m×82+12mv2mv2
=(32+12=(32+12mv2)Jmv2)J
32m+1232m+12mv2=50mmv2=50m
v2=36v2=36
v=6v=6m/sm/s
Since we see that 102=82+62102=82+62
The two balls must be moving at right angles to each other.
KE before collision =12=12m.102+0m.102+0
=50mJ=50mJ
KE after collision =12=12m×82+12m×82+12mv2mv2
=(32+12=(32+12mv2)Jmv2)J
32m+1232m+12mv2=50mmv2=50m
v2=36v2=36
v=6v=6m/sm/s
Since we see that 102=82+62102=82+62
The two balls must be moving at right angles to each other.
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12
SOLUTION:-
given by :-
》heavier ball's mass (M) = 5KG
》and intial velocity (U) = 3m/s
》and final velocity(V) = ?
》lighter ball's mass(m) = 2kg
》lighter ball's intial velocity (u) = 0
》and final velocity(v) = 2m/s
☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
》by formula conservation of momentum:-
》=>MV + mv = MU + mu
》=>5×V + 2×2 = 5×3 + 2×0
》=>5V = 15-4
》=> V = 11/5
》=> V = 2.2 m/s
》hence , after collision the velocity of the 》heavier ball's = 2.2 m/s
■I HOPE ITS HELP■
given by :-
》heavier ball's mass (M) = 5KG
》and intial velocity (U) = 3m/s
》and final velocity(V) = ?
》lighter ball's mass(m) = 2kg
》lighter ball's intial velocity (u) = 0
》and final velocity(v) = 2m/s
☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
》by formula conservation of momentum:-
》=>MV + mv = MU + mu
》=>5×V + 2×2 = 5×3 + 2×0
》=>5V = 15-4
》=> V = 11/5
》=> V = 2.2 m/s
》hence , after collision the velocity of the 》heavier ball's = 2.2 m/s
■I HOPE ITS HELP■
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