Question

The same force acts on two bodies of different masses of 9kg and 25kg initially at rest. the ratio of the times reqiured to attain the same final velocity is

Answer 1

Answer:

Let the force applied = F

Let the final velocity = v

Time taken for mass 9 kg (M1) = t

Time taken for mass 25 kg (M2) = T

Now, Impulse = Change in momentum

so, Impulse for M1 = M1v - M1 (0) = 9v - 9(0) = 9v

also, Impulse = force × time = Ft

So, Ft = 9v or F = 9v/t (eq.1)

For M2,

Impulse = M2V - M2(0) = 25v - 25×) = 25v

Also, Impulse = F×T

so, FT = 25v

or F = 25v/T (eq.2)

From equations 1 & 2,

9v/t = 25v/T

or, T/t = 25v/9v = 25/9

Hence the ratio of times taken by body of mass 25kg to that taken by the body of mass 9 kg to attain the same velocity = 25:9

Answer 2

Answer: 9:25

Explanation:

Mass of first body = 9kg

Mass of second body = 25 kg

time for first body to attain final velocity = $t_{1}$

time for first body to attain final velocity = $t_{2}$

We know that F=ma

By this for first body,

F = 9$a_{1}$          ∴$a_{1}$  = F/9

For the second body,

F = 25$a_{2}$          ∴$a_{2}$ = F/25

By   v = u + at,

$v_{1}$ = 0 + (F/9)( $t_{1}$)

$v_{2}$ = 0 + (F/25)($t_{2}$)

We know that $v_{1}$ = $v_{2}$

hence, (F/9)( $t_{1}$) =  (F/25)($t_{2}$)

∴ $t_{1}$/$t_{2}$ = $\frac{F * 9}{25 * F}$ = 9/25

Answer is 9:25

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