The speed of the block at point C, immediately before it leaves 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 C, immediately before it leaves the second incline is

Smart Achievers · Accepted Answer

(2)

$In ∆ B C E, \tan 30 ° = \frac{B E}{C E} \Rightarrow \frac{1}{\sqrt{3}} = \frac{B E}{3 \sqrt{3}} \Rightarrow B E = 3 m$

From mechanical energy conservation principle,

Mechanical energy at B = mechanical energy at C

$\frac{1}{2} M {(\sqrt{45})}^{2} + M \times 10 \times 3 = \frac{1}{2} M v_{c}^{2}$

$45 + 60 = v_{c}^{2}$ $∴$ $v_{c} = \sqrt{105} m/s$

Solution

1 $\sqrt{120} m/s$

2 $\sqrt{105} m/s$

3 $\sqrt{90} m/s$

4 $\sqrt{75} m/s$

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Solution

1 120 m/s

2 105 m/s

3 90 m/s

4 75 m/s

Ans. (2) In ∆BCE, tan30°=BECE⇒13=BE33⇒BE=3 m From mechanical energy conservation principle, Mechanical energy at B = mechanical energy at C 12M(45)2+M×10×3=12Mvc2 45+60=vc2 ∴ vc=105 m/s

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1 $\sqrt{120} m/s$

2 $\sqrt{105} m/s$

3 $\sqrt{90} m/s$

4 $\sqrt{75} m/s$