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MHT CET202016 Oct 2020Morning ShiftPhysicsMotion in One DimensionActual

A vehicle of mass 'M' is moving with momentum 'P' on a rough horizontal road. The coefficient of friction between the tyres and the horizontal road is ' '. The stopping distance is (g = acceleration due to gravity)

Options

  1. AP ² 2 g
  2. BP ² 2 gM ²
  3. CP ² gM ²
  4. DP ² 2 m ²

Correct answer

B. P ² 2 gM ²

Step-by-step solution

Initial velocity u = p m Final velocity v =0 (as the vehicle must stop) Force of friction = mg (where g is acceleration due to gravity) Acceleration due to friction =- mg m =- g (-ve sign shows that it is retardation ) Using the kinematic expression v ²= u ²=2 as and inserting various values we get stopping distance s array l (0)²- p ² ~m ² =2(- g ) s s = p ² 2 ² g array

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