A constant force is exerted on a cart that is initially at rest on a frictionles

Question

A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time interval and gives the cart a final speed. To reach the same speed using a force that is half as big, the force must be exerted for a time interval that is four times as long twice as long the same length half as long a quarter as long Two carts-A and B-on frictionless air tracks are initially at rest. Cart A is twice as massive as cart B. Now you exert the same constant force on both carts for 1 second. One second later, the momentum of cart A is twice the momentum of cart B the same as the momentum of cart B half the momentum of cart B not enough information to determine Two identical carts-A and B-initially are moving on frictionless air tracks. The initial speed of cart A is twice that of cart B. You then exert the same constant force on the two carts over 1 second. One second later, the change in momentum of cart A is non-zero and twice the change in momentum of cart B non-zero and the same as the change m momentum of cart B zero non-zero and half the change in momentum of cart B not enough information to determine

Explanation / Answer

here,

the final speed , v = u + a * t

as the initial speed is zero

so the final velocity is directly proportional to the accelration

when the force is halved

the time taken is doubled

the time taken is twice as long

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