Question

A bullet of mass 50g moving with a speed of 500m/s is brought to rest in 0.01s. Find the impulse and the average force.

Answer 1

The impulse is 25 kg-m/s, and the average force is 2500 N.

Impulse is given by the formula

J=F_avg*t=m*Δv

Plugging the values in the above equation

J=(0.05)*(500-0)=25 kg-m/s

Now the average force is calculated as

J=F_avg*t

25=F_avg*0.01

F_avg=2500 N

Therefore, The impulse is 25 kg-m/s, and the average force is 2500 N

Answer 2

Answer:Given that,

                        m=50g

                           =0.05kg

                        u=500m/s

                        v=0m/s

                        t=0.1sec.

  We know that,

             Impulse=m(v-u)

                          =0.05(0-500)

                          =0.05×(-500)

                          = -25 kg m/s

 Again,

        Average force=Impulse/time

                                 = -25/0.1

                                 = -250N

Therefore,

      The impulse is -25 kg m/s

    And the average force is -250N

Explanation:Forces are never come in negative singn . But this answers are come in negative, because resistive or retarding forces are always come in negative.

For this reason this answers have come in negative

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