Motion of masses connected by a string over a pulley

  • Thread starter songoku
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  • #26
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I am really sorry for late reply.

Let me try again:
(iv)
The two particles will experience same impulses.
Impulse A = m . Δva = [tex] m . (-v + \sqrt{\frac{2gh}{3}} ) [/tex]

Impulse B = 2m . Δvb = 2m . (v - 0)

Equating them:
[tex] m . (-v + \sqrt{\frac{2gh}{3}} ) = 2mv [/tex]

[tex] v = \frac{1}{3} \sqrt{\frac{2gh}{3}} [/tex]

Is this correct?

Thanks
 
  • #27
haruspex
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I am really sorry for late reply.

Let me try again:
(iv)
The two particles will experience same impulses.
Impulse A = m . Δva = [tex] m . (-v + \sqrt{\frac{2gh}{3}} ) [/tex]

Impulse B = 2m . Δvb = 2m . (v - 0)

Equating them:
[tex] m . (-v + \sqrt{\frac{2gh}{3}} ) = 2mv [/tex]

[tex] v = \frac{1}{3} \sqrt{\frac{2gh}{3}} [/tex]

Is this correct?

Thanks
Looks good to me.
 
  • #28
Yes, conservation of momentum is to be applied to situation in which ball B is suddenly jerked upward. In this particular case, the absolute value of the momentum of ball A prior to the jerk will equal the sum of the absolute values of the momentum of ball A and B after the jerk. Absolute value is taken because of the effect of the pulley. Note that each time ball B hits the ground, both momentum and energy is lost to the system. The velocities of the two balls will be equal and opposite after the jerk. The string is inelastic - at the time of the jerk a very sharp force is applied by the string such that the impulse delta-F times delta-t is applied to both balls and give the change in momentum of each ball. The lost kinetic energy will appear as deformations and vibrations in the two balls.

Note also that the acceleration you calculate in i is incorrect - check your math.
 
  • #29
haruspex
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The string is inelastic
It is given as inextensible.
Several on this thread had a separate conversation and concluded that the two concepts are independent. But if there is any elasticity then the problem becomes much more difficult, so the question setter probably meant it to be taken to be inelastic as well.
the acceleration you calculate in i is incorrect
Looks right to me. A net force of mg, a total mass of 3m, so an acceleration of g/3.
 
  • #30
Duh, you'r right, I was equating forces on the two masses should be equal and opposite, versus accelerations on the two masses being equal and opposite. I got a little too sophisticated in my analysis.
 
  • #32
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ii) Impulse on ground when mass B impacts: 2mΔv ?

Yes, sorry. The mass should be 2m.


For (v) :
Distance of B = twice the distance when it accelerates upwards + decelerate to stop (when A hits the ground)

Speed of B when A hits the ground
[tex]v^2 = (\frac{1}{3} \sqrt{\frac{2gh}{3}})^2 + 2 . \frac{g}{3} . 2h [/tex]
[tex] v^2 = \frac{38gh}{27} [/tex]

Distance travelled by B when decelerating:
[tex]v^2 = u^2 - 2gd[/tex]
[tex] 0 = \frac{38gh}{27} - 2gd [/tex]
[tex] d = \frac{19h}{27} [/tex]

So total distance
= [tex] 4h + 2 . \frac{19h}{27} [/tex]
= [tex] \frac{146h}{27} [/tex]

For (vi), I think I need to use formula of sum to infinity of geometric progression but I can not find the ratio. The distance travelled by B:
a. when released from rest until hits the ground = h
b. between first hit and second hit = [tex]\frac{146h}{27}[/tex]
c. between second hit and third hit (repeating the same process as (v)) = [tex]\frac{1298h}{243}[/tex]

How to find the total distance?

Thanks
 
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