Question

a man of mass 60kg is standing on a boat of mass 140kg which is at rest in still water.The man initally at 20m from the shore he started walking on the boat for 4s with constant speed 1.5m/s towards the shore.The final distance of the man from the shore is......??​

Answer 1

$\bigstar\huge\boxed{\mathtt{\red{Solution:}}}$

As per the given question ,

mass of man = 60 kg

mass of the boat = 140 kg

Distance of the man from the shore = 20 m

Then the man started to walk on the boat,

time = 4 s

speed = 1.5 m/s

Distance traveled by the man will be given by the formula,

$\implies\mathtt{s =v\times t}$

$\implies\mathtt{s = 1.5 \times 4}$

$\implies\mathtt{s = 6 m }$

Now , we know that in order to compensate   the motion of man in forward direction the boat will will move in backward direction so that the center of mass of the entire system(boat+man) remains the same

In order to find the velocity of the boat in backward direction let us apply law of conservation of momentum ,

we know that,

$\implies\mathtt{P_i=P_f}$

$\implies\mathtt{m_{man} \times v_{man}=(m_{man}+m_{boat})\times v_{system}}$

$\implies\mathtt{60 \times 1.5=(60+140)\times v_{system}}$

$\implies\mathtt{90=200\times v_{system}}$

$\implies\mathtt{ v_{system}=\dfrac{90}{200} }$

$\implies\mathtt{ v_{system}=0.45 \dfrac{m}{s} }$

Now we need to find the distance s 1 moved by the boat in backward direction

we know that,

$\implies\mathtt{s_1 =v_{system}\times t}$

$\implies\mathtt{s_1 =0.45\times 4}$

$\implies\mathtt{s_1 =1.8m}$

Final distance of the man from the shore

= total distance of the man from the shore

+ distance moved by the boat away from the shore

- distance covered by the man towards the shore

= 20 + 1.8 - 6

= 15.8 m

The final distance of the man from the shore is 15 .8 m

Answer 2

${ \bf{ \underline{ \red{ \underline{ \red{Question}}}} : - }}$

• A man of mass 60kg is standing on a boat of mass 140kg which is at rest in still water. The man initally at 20m from the shore he started walking on the boat for 4s with constant speed 1.5m/s towards the shore. The final distance of the man from the shore is ?

${ \bf{ \underline{ \red{ \underline{ \red{Solution}}}} : - }}$

{ According to question }

Let, mass of man be " m " & mass of boat be " M " .

Hence, m = 60kg & M = 140kg

• We know man initally at 20m from the shore he started walking on the boat for 4s with constant speed 1.5m/s towards the shore.

• Let, speed ( v ) = 1.5 m/s &
• time ( t ) = 4 sec

To find distance travel by man during this time period :-

Let, distance travel by man be " S "

Hence,

• S = v × t
• S = 1.5 × 4
• S = 6 m

so, distance travel by man = 6 m

• To find relative displacement of boat we use formula :-

${ \sf{ \dashrightarrow{ let \: relative \: displacement \: of}}}$

${ \sf{boat \: will \: be \: {\red{R _{s(boat)}}}}.}$

${{ \sf{\dashrightarrow{ \red{R _{s(boat)}}}}} = { \sf{ \frac{m \: \times \: S }{M \: + m} }}}$

${{ \sf{\dashrightarrow{ \red{R _{s(boat)}}}}} = { \sf{ \frac{60 \: \times \: 6 }{140 \: + \: 60} }}}$

${{ \sf{\dashrightarrow{ \red{R _{s(boat)}}}}} = { \sf{ \cancel{ \frac{360}{200}} = \frac{9}{5} = 1.8 \: m}}}$

Relative displacement of boat is

1.8 m

• To find the final distance of the man from the shore

Let, " u " will be the initial distance of man to shore { man initally at 20m from the shore }

Hence, u = 20 m & Let " x " be the final distance of the man from the shore.

${ \dashrightarrow{ \sf{{ \red{x}} = {u \: + \: {R _{s(boat)} \: - \:S }}}}}$

${ \dashrightarrow{ \sf{{ \red{x}} = {20 + 1.8 - 6}}}}$

${ \dashrightarrow{ \sf{{ \red{x}} = {20 - 4.2}}}}$

${ \dashrightarrow{ \sf{{ \red{x}} = {15.8 \: m}}}}$

Hence, the final distance of the man from the shore is 15.8 m

i hope it helps you.