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Physics: Post your doubts here!

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I found it in a topical past papers book. The year was not written.
:/ I don't think A shud be the answer...
Can u show me ur complete solution please or see mine.. tell me anything is wrong with it :s
 

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:/ I don't think A shud be the answer...
Can u show me ur complete solution please

KE Trolley(1/2 * m * v^2) + KE Mass = Loss in GPE (mgh)
1/2 x 1.5 x v^2 + 1/2 x 0.5 x v^2 = 0.5 x 9.81 x 1.0
v^2 = 4.905

KE Trolley = 1/2 x m x v^2 = 1/2 x 1.5 x 4.905 = 3.7 J

Note: Velocity of both objects is same.
 
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Oh yeahh, I was adding up their K.E's, lol -.-.
Thanks a lot
KE Trolley(1/2 * m * v^2) + KE Mass = Loss in GPE (mgh)
1/2 x 1.5 x v^2 + 1/2 x 0.5 x v^2 = 0.5 x 9.81 x 1.0
v^2 = 4.905

KE Trolley = 1/2 x m x v^2 = 1/2 x 1.5 x 4.905 = 3.7 J

Note: Velocity of both objects is same.
 
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A car is travelling with uniform acceleration along a straight road. The road has marker posts every 100m. When the car passes one post , it has a speed of 10m/s and , when it passes the next one , its speed is 20m/s

What is the car acceleration ?
A.0.6m/s^2
B.1.5m/s^2
C.2.5m/s^2
D.6.0m/s^2

(from marking scheme answer is (B:1.5m/s^2)[June 2004 paper 1 number 7]
 
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A car is travelling with uniform acceleration along a straight road. The road has marker posts every 100m. When the car passes one post , it has a speed of 10m/s and , when it passes the next one , its speed is 20m/s

What is the car acceleration ?
A.0.6m/s^2
B.1.5m/s^2
C.2.5m/s^2
D.6.0m/s^2

(from marking scheme answer is (B:1.5m/s^2)[June 2004 paper 1 number 7]
u = 10 , v = 20 , s = 100
using v^2 = u^2 + 2as
a = 1.5
 
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