The speed of the block at point B immediately after it strikes the second incline is -- [2008]

Question

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60 °$ to $30 °$ at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic ( $g = 10 {m/s}^{2}$ ). [2008]

Q. The speed of the block at point B immediately after it strikes the second incline is --

Smart Achievers · Accepted Answer

(2)

As the inclined plane is frictionless,

$The K.E. at B = P.E. at A$

$\frac{1}{2} m v^{2} = m g h \Rightarrow v = \sqrt{2 g h}$

$In ∆ A D B, \tan 60 ° = \frac{h}{\sqrt{3}} \Rightarrow h = 3 m$

$∴$ $v = \sqrt{6 g} = \sqrt{60} m/s$

This is the velocity of the block just before collision. This velocity makes an angle of $30 °$ with the vertical. Also in right-angled triangle $B E C, ∠ E B C = 60 °$ . Therefore $v$ makes an angle of $30 °$ with the second inclined plane BC. The component of $v$ along BC is $v \cos 30 °$ .

It is given that the collision at B is perfectly inelastic, therefore the impact forces act normal to the plane such that the vertical component of velocity becomes zero. The component of velocity along the incline BC remains unchanged and is equal to $v \cos 30 °$

$= \sqrt{60} \times \frac{\sqrt{3}}{2} = \sqrt{\frac{180}{4}} = \sqrt{45} m/s$

Solution

1 $\sqrt{60} m/s$

2 $\sqrt{45} m/s$

3 $\sqrt{30} m/s$

4 $\sqrt{15} m/s$

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Solution

1 60 m/s

2 45 m/s

3 30 m/s

4 15 m/s

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1 $\sqrt{60} m/s$

2 $\sqrt{45} m/s$

3 $\sqrt{30} m/s$

4 $\sqrt{15} m/s$