Two men A and B of masses M and M+m respectively start simultaneously from the ground and climb with uniform acceleration up the free ends of massless inextensible rope which passes over a smooth pulley at a height h from the ground. Which man will reach the pulley first??
a) A
b)B
c)Both reach simultaneously
d)Data insufficient
11
The ring R in the arrangement shown can slide along a fixed horizontal rod XY. It is attached to the block B by a light string. The block is released from rest with the string horizontal.
a)One point in the string will have only vertical motion.
b)R and B will always have moments of the same magnitude.
c)When the string becomes vertical , the speeds of R and B will be inversely proportional to their masses.
d) R will lose contact with the rod at some point.
(please explain in detail)
1for the second question i think its (a) and the point is COM as it as zero displacement in horizontal dirn (no net Fext)
1i posted the answers before u posted the answers the explanation is as fillows
a) simple let the masses of both be m and the length is l therefore both the masses will move l/2 and the strng that was intially at pnt l/2 will not move at all...
c) since force in horizonatal direction is 0 therefore the total momentum in the horizontal direction is 0
mv=Mv therefore u get it
b) i didnt know that bcz the word moments of magnitude hv no apparent meaning in this question
d) the tension force always acts in the downwards direction for r therefore T+mg=N
terefore N can never be 0 and the object will never loose contact
113) A block is placed at the bottom of an inclined plane and projected with some initial speed. It slides up the incline , stops after time t1 and slides back in a further time t2. The angle of inclination of the plane with the horizontal is theta. and the co-efficient of friction is \mu.
a) t1 > t2
b) t1 < t2
the answer given is b.is the answer correct...shud'nt it be a
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4) Two blocks A and B of same masses are joined by a light string and placed on a horizontal surface. An external horizontal force P acts on A. The tension in the string is T.The force of friction on A and B are F1 and F2 respectively.The limiting value of F1 and F2 is Fo.As P is gradually increased..
a) for P < Fo , T = 0
b)for Fo < P < 2Fo , T = P = -Fo
c)for P > 2Fo , T=P/2
d) None of these.
answer given ABC (please explain)
11Thank you Varun.tinkle...
someone please try the first one also..(explanation)
11st asnwers very simple the conceptual part of it i ma leaving it for u 2 think
T-(M+m)g=ma
T-Mg=ma t will be same on both sides
therfore the one with the less mass will reach 1st that is A
13) i think u are wrong can anyone pls confirm it...
my working in these situations i assume @ and u to make calc. easier so
u=0.1 g=10 @=45 and v=10
therfore
t1=10/(10/√2 +1/√2)
and distance before it stops
100/2(10/√2 +1/√2)
therefore time t22
100/((10/√2 +1/√2)(10/√2 -1/√2))
therefore t= approx is 10/7 and therefore t2>t1
14) its simple
first
F-F_{0}-T=ma
and
T-F_{0}=ma
therefore
F>2F_{0}
now
now if F>F_{0}
then T has to be 0 otherwise
similarly u cna solve ofr the other cases