A car and a motorcycle are moving with the same momentum. When equal retarding forces are applied, the car comes to halt in $t_{1}$ seconds and the motorcycle in $t_{2}$ seconds. If the mass of the car is five times more than the mass of the motorcycle, then

t_{1} = t_{2}

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t_{1} = 5 t_{2}

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t_{1} = \frac{1}{5} t_{2}

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t_{1} = 25 t_{2}

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Solution

The correct option is C $t_{1} = t_{2}$
As we know force is directly proportional to the rate of change of momentum.here momentum and retarding force is same.so time taken will be same.
$F \propto \frac{d p}{d t}$
hence, $t_{1} = t_{2}$

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A car and a motorcycle are moving with the same momentum. When equal retarding forces are applied, the car comes to halt in t1 seconds and the motorcycle in t2 seconds. If the mass of the car is five times more than the mass of the motorcycle, then

The correct option is C t1=t2As we know force is directly proportional to the rate of change of momentum.here momentum and retarding force is same.so time taken will be same.F∝dpdthence,t1=t2

A car and a motorcycle are moving with the same momentum. When equal retarding forces are applied, the car comes to halt in $t_{1}$ seconds and the motorcycle in $t_{2}$ seconds. If the mass of the car is five times more than the mass of the motorcycle, then

The correct option is C $t_{1} = t_{2}$
As we know force is directly proportional to the rate of change of momentum.here momentum and retarding force is same.so time taken will be same.
$F \propto \frac{d p}{d t}$
hence, $t_{1} = t_{2}$