95 - Kinematics Questions Answers

A cylindrical pipe of radius r is rolling towards a frog sitting on the horizontal ground. Center of the pipe is moving with velocity v. To save itself ,the frog jumps off and passes over the pipe touching it only at top. Denoting air time of frog by T, horizontal range of the jump by R and acceleration due to gravity by g. Find T and R.

Asked By: MUDIT
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Joshi sir comment

According to given condition 2r = uy2/2g, T= 2uy/g =24gr/g=4r/g At top point due to touch, vecocity of the frog be ux2v. R=[uxuy/g]+[ux2v)uy/g]=[uy/g](2ux2v) =[2uy/g](uxv) =T(uxv) Now minimum possible value of velocity of frog at top to  cross the cylinder is gr (By the concept of vertical circle) now solve carefully without error

A small ball is thrown from foot of a wall with minimum possible velocity to hit a bulb B on the ground at a distance L away from the wall.Find the expression for height h of shadow of the ball on the wall as a function of time t.Acceleration due to gravity is g.

Asked By: MUDIT
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Joshi sir comment

While making this paper, I stayed up till late at night working with candle light. I had two

candles of equal length L lit and placed at the same time, as shown in the figure. Immediately

thereafter I noticed that the shadow of the first candle on the left wall is not changing in

length, and the shadow of the second candle on the right wall is being shortened at a speed v.

(Take d1

= 10 cm, d2

= 20 cm, d3

= 30 cm, v = 1 mm/s, L = 20 cm.)

Asked By: AAYUSH MALAVIYA
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Joshi sir comment

Problem is interesting yet not complete. Complete it to get answer 

thanks

Two boys enter a running escalator at the ground floor of a shopping mall . The first boy repeatedly follows a cycle of p1=1 step up and then q1=2 steps down whereas the second boy repeatedly follows a cycle of p2=2 steps up and then q2=1 step down. Both of them move relative to escalator with a speed v=50cm/s . If the boys take t1=250s and t2=50s respectively to reach the first floor in complete number of cycles, how fast is the escalator running ? 

 

Asked By: JATIN KUMAR
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Joshi sir comment

let u and v are the speeds of boy and escalator according to given condition  [1(u+v)T-2(u-v)T]n1=[2(u+v)T-1(u-v)T]n2  [1(u+v)-2(u-v)]Tn1=[2(u+v)-1(u-v)]Tn2 Given that Tn1=250 and Tn2=50 [1(u+v)-2(u-v)]250=[2(u+v)-1(u-v)]50 now solve it for v, u is given 

A particle projected from the ground passes two points , which are at height 12m and 18m above the ground and a distance d=10m apart . What could be the minimum speed of projection. Acceleration due to gravity is g=10m/s^2 . 

Asked By: JATIN KUMAR
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Joshi sir comment
Sir along with the written question, can you also calculate the velocity of the midpoint of CD( marked in green) under the same conditions as in question? Thank you
Asked By: DARIUS
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Joshi sir comment

it

Asked By: FURQAAN
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Joshi sir comment
Physics >> Mechanics part 1 >> Kinematics Medical Exam

the min no. of non coplanar and non zero vectors of unequal magnitude which can give zero resultant is _________??

Asked By: SARIKA
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Joshi sir comment

four

let three are i, j, k then fourth will be -(i+j+k)

Physics >> Mechanics part 1 >> Kinematics Medical Exam

if y= 50 / [(x-5)2 +5]2  then maximum value of y will be??     

      

Asked By: SARIKA
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Joshi sir comment

let z = [(x-5)2 +5]

then dz/dx = 2(x-5)

compare it to 0 we get x = 5

so for x = 5, z will be minimum and so y will be max for that x

put x= 5 and get answer 

A particle has initial velocity of y = 3i+ 4j and a constant force F = 4i -3j acts on it. Then what will be the path followed by the particle?

Asked By: VIKRAM
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Joshi sir comment

acceleration is constant and initial velocity is perpendicular to it because dot product of two given vectors is 0 so this is the case equivalent to a particle thriwn from the top of a building horizontally under gravity so path will be parabolic

if you want the equation of path then reply

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