Question

A small block of mass mi is released from rest at the top of a curve-shaped, frictionless wedge which sits on a frictionless horizontal surface as shown. The height of the wedge is h = 5 m. When the block leaves the wedge its velocity relative to the ground is measured to be 4.00 m/s to the right as shown in the figure. If the mass of the block is doubled to become 2m, what will be its speed when it leaves the wedge? mo 4,00 m/s (a) (b)

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Answer 1

Solution:

Net External Force = Gravity in Y-Direcion

Total Momentum is conserved in X-Direction.

i.e.,

Total Mechanical Energy is Conserved.

i.e.,

migh = mivi +։mov — — — – — — — — - (Eq – 2)

From Eq -1

$v_{2} =\frac{m_{1}v_{1}}{m_{2}}$

$m_{1}gh=\frac{1}{2}m_{1} v_{1}^{2} + \frac{1}{2} \frac{m_{1}^{2} v_{1}^{2}}{m_{2}}$

Therefore we get,

$v_{1}^{2}= \frac{2gh}{1+\frac{m_{1}}{m_{2}}}$

when $v_{1}= 4m/s => 16 = \frac{2*10*5}{1+\frac{m_{1}}{m_{2}}}$

$\frac{m_{1}}{m_{2}}=5.25$

Now When the mass is doubled, we get

$\frac{m_{1}}{m_{2}}=10.50$

$v_{1}=\sqrt{\frac{2*10*5}{1+10.5}}=2.95 m/s$

Therefore the 2.92 m/s is the speed when it leaves the wedge.