# Finding the value of velocity $u$ [closed]

A block of mass $m$ is pushed towards a moveable wedge of mass $\eta m$ and height $h$, with a velocity $u$. All surfaces are smooth. The minimum value of $u$ for which the block will reach the top of the wedge is:

1. $\sqrt{2gh}$
2. $\eta \sqrt{2gh}$
3. $\sqrt{2gh(1+\frac1\eta)}$
4. $\sqrt{2gh(1-\frac1\eta)}$

What's confusing me is how to proceed knowing the wedge is moveable. I tried drawing the FBD of both the objects. But I am not getting any far with it.

## 2 Answers

Firstly, as the wedge is movable, it will move when the object slides onto it (I guess that's what they meant). The steps you may follow:

1. momentum is conserved.So, the initial momentum of the system will be $mu$. The final momentum will be the combined momentum of the wedge and the block $(m+m*eta)v$. From here, you can find an expression for the final velocity in terms of u

2.Using conservation law of energy, $$K.E_i = K.E_f+mgh$$ 3.By solving this, you can solve for minimum value of u

I don't understand what you mean by a movable wedge but i am sure that we need the angle the wedge makes with the horizon and then you calculate the force and deceleration on the box moving towards the top and finally you will get the minimum velocity for it to get up the wedge?

perhaps i didn't understand the question please explain it more?

Because all i can think of now is when you push the box some of the kinetic energy will go to the movable wedge and it will start moving...especially since all the surfaces are smooth.

There is one more thing i can think of and that is conservation of energy ...and by that i mean that the kinetic energy and weight(it changes with height) when it is down the wedge should equal the one when it is on top and that means that it should be stationary on top and that means the (weight on top)-(weight on bottom)=kinetic and from there its easy to get the speed...

You can try this the kinetic energy of the wedge and the box at that height is equal to the kinetic energy of the box movement and with the coservation of momentum you get the movement speed of the wedge when the box hits it and then you can calculate the box's initial speed?

I think the last one is the solution and I hope I helped!

• Well, I have not been given the angle and neither does it appear in any of the options. But for calculations, we can assume it to be $\theta$. Moreover, this is surely a problem of conservation of energy, and I tried this method. But then, how does $\eta$ come into the picture? How do I consider the motion of the wedge? – Tejas Jan 3 '14 at 12:41
• what is Î· exactly? – Fahadalkadhi95 Jan 3 '14 at 12:42
• I assume it's a constant. Like the mass of the wedge is '$\eta$' times the mass of the block. – Tejas Jan 3 '14 at 12:43
• I googled that and it is energy conversion efficiency ? – Fahadalkadhi95 Jan 3 '14 at 12:46
• and also Î·m is the mass of the wedge? – Fahadalkadhi95 Jan 3 '14 at 12:47