Question

A person of mass 60kg is standing on a raft of mass 40kg in a lake. the distance of the person from the bank is 30m. if the person starts running towards the bank with velocity 10m/s, th n what will his distance be from the bank after one second ?​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of man (M) = 60 Kg.
• Mass of person (m) = 40 Kg.
• Man starts with a velocity (u) = 10 m/s. [Final]
• And, the velocity of boat (v) = v [Final]

$\huge{\bold{\underline{Explanation:-}}}$

Here we have to use Law of conservation of momentum,

From, the formula,

$\large{Mu_1 + mu_2 = Mv_1 + mv_2}$

Initial momentum is zero as both are at rest.

Now,

$\large{Mu_1 + mu_2 = 0}$

Now,

$\large{u_1 = u = 10 \: m/s}$

$\large{u_2 = v}$

Substituting the values,

$\large{Mu + mv = 0}$

Substituting the values,

$\large{60 \times 10 + 40 \times v = 0}$

$\large{600 + 40v = 0}$

$\large{40v = - 600}$

$\large{v = \dfrac{-600}{40}}$

$\large{v = - 15 \: m/s}$

Negative Sign indicates the velocity is opposite to the motion of the man.

$\rule{300}{1.5}$

Finding the relative velocity of man w.r.t ground.

$\large{ \tt v_r = v_m + v}$

Substituting the values,

$\large{ v_r = 10 + (-15)}$

$\large{v_r = 10 - 15}$

$\large{\boxed{ \tt v_r = - 5 \: m/s}}$

$\rule{300}{1.5}$

Distance travelled in 1 seconds.

$\large{\bold{ \tt Distance = Speed \times time}}$

Substituting the values,

$\large{ Distance = - 5 \times 1}$

$\large{\boxed{ \tt Distance = - 5 \: meters}}$

$\rule{300}{1.5}$

Final distance,

$\large{Distance \: away = 30 - Distance \: travelled}$

Now,

$\large{Distance \: away = 30 - (-5)}$

$\large{Distance \: away = 30 + 5}$

$\huge{\boxed{\boxed{ \tt Distance \: away = 35 \: meters}}}$

So, the distance away from the bank is 35 meters.

Note:-

• In the above case when distance comes negative it specifies that the resultant distance is backwards but not forward.