A lorry and a car moving with the same momentum are brought to rest by the application of brakes which provide equal retarding forces,

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Solution

a) Here, work done is provided by kinetic energy

$K E 1 = F 1 \times d 1$ ...(1)

$K E 2 = F 2 \times d 2$ ...(2)

Also, $K E 1 = K E 2$ and $F 1 = F 2$ ,

$d 1 = d 2$

Both vehicles will stop at same distances

b) Also $m 1 u 1$ = $m 2 u 2$

$m 1 \times \frac{d 1}{t 1} = m 2 \times \frac{d 2}{t 2}$

But, $d 1 = d 2$ thus
$\frac{t 1}{t 2} = \frac{m 1}{m 2}$

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A lorry and a car moving with the same momentum are brought to rest by the application of brakes which provide equal retarding forces,

a) Here, work done is provided by kinetic energy KE1=F1×d1 ...(1) KE2=F2×d2 ...(2) Also, KE1=KE2 and F1=F2, d1=d2 Both vehicles will stop at same distances b) Also m1u1 = m2u2 m1×d1t1=m2×d2t2 But, d1=d2 thus t1t2=m1m2