Recent content by Shivam

Bro , i thought geometrically and angle was not equal to alpha , i also tried this [ Tan(theta) = Cot(alpha), Tan(theta) = Tan(90-alphal), So, theta = 90 - alpha], is it correct

After impact VCos(alpha) will be normal to the inclined, now i calculated for theta by using [ Tan(theta) = vCos(alpha)/vSin(alpha),. Tan(theta)= Cot(alpha) ]. I don't know how to solve further to get value of theta, according to book thta = alpha.

Thanx for the Welcome, and you were right, the actual value of speed was 9.33333333 and i got the right answer using that.

I know how to solve along x and y-axis but i can't think of how to start solving in the dricection on m1.

My attempt- a) used equation { MV + mv = (M+m) V' } and got the right answer. b) I assumed that potential energy was asked for when the two balls were moving together with velocity 9.3m/s, so I used that when before the collision K.E( of m) + K.E( of M) will be equal to K.E(of M+m) +P.E(...

Thanks for the reply.. I think the question just wanted to know value of frictional force.

1). I calculated maximum safe velocity using the equation - V(max)=√200x10x0.2 =20m/s So the speed at which car is traveling is greater than the safe speed.. So the car should skid. So why 4th option is not correct ?

First i am sorry for writing wrong Units and thanks for the torque units and also for telling me about the tension will be variable.. can you lastly tell me how tension will have zero work as internally, if you explainna little bit..

My attempt at the solution... I only have problem in solving part a) 1) i calculated the spring force actingbon the table block and it is greater than the frictionnand tension force acting in the opposite direction, so by that the block will move in left. 2) now i found that which forces are...

Can you please explain the answer, I have been stuck with this for 3 days now, 2 days I thought about it myself and then yesterday I posted this question

Let's think of this, even if N1 <mgCos(theta) then let's make it's components one will be in horizontal and one will be in vertical and the vertical component of N1 will be canceled by the Ng given by the ground, then what will cancel horizontal component. ?

What to do next I can't figure it out...

ok, it won't be, let me try drawing FBD.

the force mass m exerts is the Normal reaction between them. The mg will have two component mgcos(theta) and mgSin(theta), the mgSin(theta) is just parallel to inclined surface so it will not affect the wedge, the mgCos(theta) is pushing on the wedge, so what's next