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mechanics

Easy

Question

$bold italic P bold italic a bold italic s bold italic s bold italic a bold italic g bold italic e bold colon A space b l o c k space o f space m a s s space 1 space k g space i s space m o v i n g space t o w a r d s space a space m o v a b l e space w e d g e space o f space m a s s space 2 space k g space a s space s h o w n space i n space f i g u r e. space A l l space s u r f a c e s space a r e space s m o o t h. space W h e n space t h e space b l o c k space l e a v e s space t h e space w e d g e f r o m space t o p comma space i t s space v e l o c i t y space i s space m a k i n g space a n space a n g l e space theta space equals space 30 º space w i t h space h o r i z o n t a l. T h e space v a l u e space o f space v subscript 0 space i n space m divided by s space i s space$

$4$
$10$
$7$
$9$

The correct answer is: $7$

$B y space c o n s e r v a t i o n space o f space m o m e n t u m space 1 cross times y subscript 2 cos 30 degree plus 2 v w equals 1 cross times v subscript 0 text Now, net velocity of block, end text f subscript b equals V subscript w plus V subscript text bw end text end subscript text also end text V subscript b text makes end text 30 degree text with horizontal and end text V subscript b w end subscript text is at end text 60 degree text with horizontal end text therefore open vertical bar v subscript w close vertical bar equals open vertical bar v subscript b w end subscript close vertical bar text as resultant end text f subscript b text passes through angle bisector. end text therefore text from equation end text left parenthesis 1 right parenthesis square root of 3 v subscript W x end subscript cross times cos 30 degree plus 2 v subscript w equals v subscript 0 7 over 2 v subscript u x end subscript equals v subscript 0 space space space space.... space left parenthesis i right parenthesis text Now, by conservation of energy, end text 1 half cross times 1 cross times v subscript b superscript 2 plus 1 half cross times 2 cross times v subscript w superscript 2 equals 1 half cross times 1 cross times v subscript 0 superscript 2 text solving, end text v subscript 0 equals 7 m divided by s$

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$A space b a l l space o f space m a s s space m space i s space m o v i n g space t o w a r d s space a space b a t s m a n space a t space a space s p e e d space v. space T h e space b a t s m a n s t r i k e s space t h e space b a l l space a n d space d e f l e c t s space i t space b y space a n space a n g l e space theta space w i t h o u t space c h a n g i n g space i t s space s p e e d. space T h e i m p u l s e space i m p a r t e d space t o space t h e space b a l l space i s space g i v e n space b y space colon$

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PhysicsMechanics

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PhysicsMechanics

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A projectile is fixed with velocity v₀ = 2m/s at angle 60º with horizontal. At top of its trajectory it explodes into three fragment of equal mass. First fragment retraces the path, second moves vertically upward with speed $fraction numerator 3 straight V subscript 0 over denominator 2 end fraction$ . The speed of third fragment (in m/s) is –

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PhysicsMechanics

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A steel ball of radius R = 20 cm and m = 2 kg is rotating about a horizontal diameter with angular speed w₀ = 50 rad/sec. This rotating ball is dropped on a rough horizontal floor and fall freely through a height h = 1.25 m. The coefficient of restitution is e = 1 and coefficient of friction between ball and floor is $mu equals 0.3$ . Then the distance in m between the point of first and second impact of the ball and floor is

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