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A cart of mass 0.5kg is placed on a smoo...

A cart of mass `0.5kg` is placed on a smooth surface and is connected by a string to a block of mass `0.2 kg.` At the initial moment the cart moves to the left along a horizontal plane at a speed of `7m//s ("Use "g=9.8m//s^(2))`

A

The acceleration of the cart is `(2g)/(7)` towards right.

B

The cart comes to momentary rest after `2.5s`.

C

The distance travelled by the cart in the first `5s` in `17.5m`.

D

The velocity of the cart after `5s` will be same as initial velocity.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`rArr0.2g=0.7a`
`rArra=(2g)/(7)m//s`
For the case, it comes to rest when `V=0`
`0=7+(0(2g)/(7))trArrt=(49)/(2g)=2.5s`

Distance travelled till it comes to rest
`0=7^(@)+2(-(2g)/(7))s`
`S=8.75m`
So in next `2.5s`, it covers `8.75m` towards right.
Total distance `=2xx8.75=17.5m`
After `5s`, it speed will be same as that of initial `(7 m//s)` but direction will be reversed.
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  • A cart with a mass M=(1)/(2) kg is connected by a string to a mass of m=200 g. At the initial moment the cart moves to the left along a horizontal plane at a speed v_(0)=7 m//s . Distance covered by smooth cart after 5 s is (g=9.8 m//s^(2))

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