The block has a mass of 40kg and rests on the surface of the cart having a mass

Question

The block has a mass of 40kg and rests on the surface of the cart having a mass of 90kg . (Figure 1) If the spring which is attached to the cart and not the block is compressed 0.2 m and the system released from rest, determine the speed of the block with respect to the cart after the spring becomes undeformed. Neglect the mass of the wheels and the spring in the calculation. Also neglect friction. Take k = 230N/m Express your answer to three significant figures and include the appropriate units

Explanation / Answer

Solution:

mass of block = m =40 kg

mass of cart =M = 90 kg

spring constant = k = 230 N/m

initial kinetic energy of block and cart = 0

The elastic potential energy due to the compression of the spring = 1/2 kx^2 = 1/2 (230)(0.2)^2 = 4.6 joules

kinetic energy of the block = 1./2* 40 *vb^2 = 20 vb^2

kinetic energy of the cart = 1/2 * 90*vc^2 = 45 vc^2

From the conservation of kinetic energy , 4.6 = 20vb^2 + 45vc ^2

and from momentum conservation, total initial momentum = total final momentum

=> 0 = 40vb - 90vc

=> 90vc= 40vb

=> vc = 40/90 vb

=> vc = 0.44vb

using this in the energy equation, and eliminating vc

4.6 = 20vb^2 + 45 (0.44vb^2)

     = 20 vb^2 +20vb^2

=>vb^2 = 4.6 /40

=> vb = 0.339 m/s

vc= 0.44 vb = 0.44*0.339 = 0.149 m/s

since the speed of cart and block are in opposite directions,

the speed of block with respect to the cart =vbc = vb-vc =(0.339)-(-0.149) = 0.488 m/s

                                                                           = 4.88 x10^-1 m/s

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