Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact). One of the bobs is released after being displaced by 10o so that it collides elastically head-on with the other bob.
(a) Describe the motion of two bobs.
(b) Draw a graph showing variation in energy of either pendulum with time, for 0≤t≤2T, where T is the period of each pendulum.
This is a short answer type question as classified in NCERT Exemplar
At t=0 suppose bob B is displaced by angle 10 to the right . it is given potential energy E1=E . energy of A, E2=0
When B is released it strikes at A at t=T/4 in the head on elastic collision between B and A comes to rest and A gets velocity of B. therefore E1=0 and E2=E. at A =2T/4, B reaches its extreme right position when KE of A is converted into PE=E2=E . Energy of B, E1=0
At t=3T/4. A reaches its mean position when its PE is converted into KE =E2 =E. it collides elastically with B and transfers whole of its energy to B. thus E2=0 and E1 =E . the entire process is r
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This is a short answer type question as classified in NCERT Exemplar
At t=0 suppose bob B is displaced by angle 10 to the right . it is given potential energy E1=E . energy of A, E2=0
When B is released it strikes at A at t=T/4 in the head on elastic collision between B and A comes to rest and A gets velocity of B. therefore E1=0 and E2=E. at A =2T/4, B reaches its extreme right position when KE of A is converted into PE=E2=E . Energy of B, E1=0
At t=3T/4. A reaches its mean position when its PE is converted into KE =E2 =E. it collides elastically with B and transfers whole of its energy to B. thus E2=0 and E1 =E . the entire process is repeated.
b)the plot of energy with time 0
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<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>At t=0 suppose bob B is displaced by angle 10 to the right . it is given potential energy E<sub>1</sub>=E . energy of A, E<sub>2</sub>=0</p><p>When B is released it strikes at A at t=T/4 in the head on elastic collision between B and A comes to rest and A gets velocity of B. therefore E<sub>1</sub>=0 and E<sub>2</sub>=E. at A =2T/4, B reaches its extreme right position when KE of A is converted into PE=E<sub>2</sub>=E . Energy of B, E<sub>1</sub>=0</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745465255phphO5SSr_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745465255phphO5SSr.jpeg" alt="" width="178" height="146"></picture></div><div><p>At t=3T/4. A reaches its mean position when its PE is converted into KE =E<sub>2</sub> =E. it collides elastically with B and transfers whole of its energy to B. thus E<sub>2</sub>=0 and E<sub>1</sub> =E . the entire process is repeated.</p><p>b)the plot of energy with time 0</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745465306php5YH8AC_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745465306php5YH8AC.jpeg" alt="" width="463" height="400"></picture></div></div></div></div>
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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