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A 3.0kg block slides on a frictionless h...

A `3.0kg` block slides on a frictionless horizontal surface, first moving to the left at `50m//s`. It collides with a spring as it moves left, compresses the spring and is brought to rest momentarily. The body continues to be accelerated to the right by the force of compressed spring. Finally, the body moves to the right at `40m//s`. The block remains in contact with the spring for `0.020s`. What were the magnitude and direction of the impulse of the spring on the block? What was the spring's average force on the block?

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Verified by Experts

The correct Answer is:
A, B, C

`Impu lse=Deltap=p_f-p_i`
`=m(v_f-v_i)`
`=(3)[40-(-50)]`
`=270N-s`
i.e. impulse is `270N-s` (towards right)
`Impu lse=F_(av)Deltat`
`:. F_(av)=(Impu lse)/(Deltat)=(270)/(0.02)`
`=13500N`
`=13.5kN` (towards right)
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