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- Homework Statement
- A small block of mass M moves on a frictionless surface on an inclined plane, as shown in the figure. The angle of the incline suddenly changes from 60 to 30 at point B. The block is initially rest at A

- Relevant Equations
- $$ \frac{mv^2}{2} = mgh$$

momentum conservation

So, what I did was suppose the mass of ramp is $ M_r$ and let velocity at B of block be v, then, after inellastic collsion both bodies v' velocity

at B ,

$$M\vec{v}= M_r \vec{v'}+ M \vec{v'}$$

or,

$$ \frac{M}{M +M_r} \vec{v}= \vec{v'}$$

Now,

Suppose I take the limit as mass of ramp goes to infinity, then I get $\vec{ v'}=0$

that means, after the collision, the block is temporarily at rest and then slides to ground.So, we can just neglect upper half and do energy conservation in lower half ramp to get velocity at final position.

However, my answer came out wrong for some reason, I wish to ask if my concepts are correct?

answer I got:

$$| v|= \sqrt{60}$$

actual answer:

$$| v|= \sqrt{105}$$

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