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Two frictionless gliders, P and Q, on a linear air track are fitted with magnets so they repel one another.
They are held near one another by a thread from one to the other, and they are positioned at rest.
Then the thread is burned through so that they move away from one another.
P moves away at a speed of 0.20 m s-1 and Q moves away at 0.12 m s-1.
Given the mass of P is 0.80 kg, calculate
(a)  the mass of Q
(b)  the initial P.E. of the system,
(c)  the speed at which P would have moved away if Q had been held fixed when the thread was burned through.
 








My answer:
mp=0.80 kg
vp = 0.20 m s-1
mq=?
vq = 0.12 m s-1.
 
(a)
mpvp - mq vq = 0
mq= mp (vp / vq) = (0.80 kg)(0.20 / 0.12)= (0.80)(20/12) kg
    = (0.80)(5/3) kg
    = 1.33 kg
 
(b)
P.E.= (1/2)(mp)(vp)2+ (1/2)(mq)(vq)2 
= (1/2)(0.80 kg)(0.20 ms-1)2 + (1/2)(5/3)(0.80 kg)(0.12 m s-1)2 
= (1/2)(0.80 kg)(0.20 ms-1)2 + (1/2)(5/3)(0.80 kg)(3/5)2 (0.12 m s-1)2 
= (1/2)(0.80 kg)(0.20 ms-1)2 {1 + (3/5)}  
= (1/2)(0.80 kg)(0.20 ms-1)2 (8/5)  
= (4/5)(0.80 kg)(0.20 ms-1)2
= 0.0256 J
 
(c)
P.E.= (1/2)(mp)(vp)2
(vp)2 = (P.E.)(2) /(mp) = (0.0256 J)(2) / (0.80 kg) = (0.0256)(2) / (0.80)
        = 0.064
vp = √(0.064) = √(640/10000)= (8√10)/(100)
    = (25.29822128134703) / (100)
    = 0.25m/s
 
 
Understanding Physics for Advanced Level(p.41, Problem 2.28)

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