Question

A 748-N man stands in the middle of a frozen pond of radius 4.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2-kg physics textbook horizontally toward the north shore at a speed of 4.0 m/s. How long does it take him to reach the south shore

Answer 1

Answer:

The man will take 64 seconds to reach to the south shore of the frozen pond.

Step-by-step explanation:

Given:

Weight of the man = 748 N      Mass of the man,$(m)= \frac{W}{g}$ = $\frac{748}{9.8}$ = $76.32$ kg

Radius of the pond $(r)$ = 4 m

Mass of the textbook = 1.2 kg

Velocity at which the textbook is thrown = 4 ms^1

We have to find the velocity of the man after the throw.

Let the velocity is $V_m$ .

Now using law of conservation of momentum we can find the $V_m$ value.

$m_(_b_)V_b_(_i_) +m_(_m_)V_m_(_i_) =m_(_b_)V_b_(_f_)+m_(_m_) V_m_(_f_)$

Considering $V_m_(_f_)=V_m$

And initial velocity of both the man and book i.e $V_b_(_i_)=0,\ V_m_(i_)=0$

So,

⇒ $0 =m_(_b_)V_b_(_f_)+m_(_m_) V_m$

⇒ Plugging the values.

⇒ $V_m=-\frac{m_(_b_)V_b_(_f_)}{m_(_m_)}$

⇒ $V_m=-\frac{1.2\times 4}{76.32}$

⇒ $V_m=-0.062$ ms^-1

Here the negative velocity is meant for opposite direction of the throw.

Numerically we will write, $V_m = 0.062$

With this velocity the man will move towards south.

We have to calculate the time taken by the man to move to its south shore.

And we know $velocity(v)\times time(t) = distance(d)$

Let the time taken be $t$ and $v\times t = d$ and $d=r$ then, $V_m\times t=r$

Then

⇒ $t=\frac{radius\ (r)}{V_m}$

⇒ Plugging the values.

⇒ $t=\frac{4}{0.062}$

⇒ $t =64$ sec

The man will take 64 seconds to reach to the south shore of the frozen pond (circular).