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Physics: Post your doubts here!

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calculte the difference of energy....
this energy would be lost due to the frictional efeects....now calcate the force by Workdone=force*d
 
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Please share the link for the examiner report for November 2011's paper if any one has it asap. Thanks in advance :p
 
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Using Principle of linear momentum,

Sum of momentum before = Sum of momentum after

Initially, they are both at rest => v = 0
Therefore the momentum before is 0

After the cannon ball is fired, the cannon ball moves forward a the cannon experiences a recoil velocity, that is, it moves backward.

Therefore, sum of momentum after,
= (1000 * - 5) + (10 * v)

Sum of momentum before = Sum of momentum after

0 = (1000 * - 5) + (10 * v)
0 = -5000 + 10v
10v = 5000
v = 500 m/s

Note : -5 is used since the cannon is experiencing a recoil force in the opposite direction.

Hope it helps.
 
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yar...firslty u have to calculate the velocity of ball S..
now apply elastic collision cases...if any of them suits here then it would be a elastic collision or otherwise inelastic..
for elastic..relatve speed of approach=speed of separation..
speed of approach=4(velocity of ball before collision)+0(velovity of ball S).
speed of separation=0.8(fromgraph)+speed of ball S after the collision)....(dont use the negative sign with value of velocity from the graph,as negative sign only sows a change in direction)...

if spped of approach turns out to be equal to speed of separation..then elastic..otherwise inelastic
 
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Umm, i want to know which papers i'm going to give for physics for my AS..
i mean in the timetable its phy paper 21,31,41,32,51,11 :\
 
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you will use the conservation of momentum i.e. final momentum - initial momentum =0
yar...firslty u have to calculate the velocity of ball S..
now apply elastic collision cases...if any of them suits here then it would be a elastic collision or otherwise inelastic..
for elastic..relatve speed of approach=speed of separation..
speed of approach=4(velocity of ball before collision)+0(velovity of ball S).
speed of separation=0.8(fromgraph)+speed of ball S after the collision)....(dont use the negative sign with value of velocity from the graph,as negative sign only sows a change in direction)...

if spped of approach turns out to be equal to speed of separation..then elastic..otherwise inelastic
Thanks!
 
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