Question

a car can be stopped over a distance x when its momentum is p. what will be the stopping distance when the momentum is 2p
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Answer 1

: A car can be stopped over a distance x when its momentum is p.what will be the stopping distance when the momentum is 2p?

One form of Newton’s 2nd Law (most commonly seen as F = ma) is F = dp/dt, or the force is the derivative of momentum with respect to time.

If you assume a constant force causing deceleration of the car, then dp/dt is constant. v = p/m, where v is velocity. The total stopping distance is then the integral of v*dt. Since F is constant, dt = dp / F, and therefore the stopping distance is the integral of (v/F)*dp = p/(m*F) *dp. The bounds of the integral will be from p to 0 in the first case, and 2p to 0 in the second case. This integral evaluates to p^2/(2*m*F) = x based on the given information for the first case, and in the second case we get (2p)^2/(2*m*F) = 4*(p^2/(2*m*F)) = 4*x in the second case. So if the initial momentum is doubled the stopping distance is multiplied by 4.

Step-by-step explanation:

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