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Butoxors [25]
3 years ago
8

A 1.5kg cart moves along a track at 0.20m/s until it hits a fixed bumper at the end of the track. Find the average force exerted

on the cart by the bumper if the cart is in contact with the bumper for 0.10s and it rebounds back along original direction with speed 0.125m/s
Physics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

Net force will be 4.875 N

Explanation:

We have given mass of the cart m = 1.5 kg

Initial velocity of the cart  = 0.2 m/sec

And final velocity of the car = - 0.125 m/sec ( Negative direction is due to opposite direction )

Instant of time \Delta t=0.1sec

Change in momentum is given by

\Delta P=1.5\times (0.2-(-0.125))=1.5\times 0.325=0.4875kgm/sec

Now force is given by

F=\frac{\Delta P}{\Delta t}=\frac{0.4875}{0.1}=4.875N

Net force will be 4.875 N

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ser-zykov [4K]

Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

                       = 8.17 N/m

The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

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Answer:

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We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

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