Question

Q15 A cat of body mass 2.5 kg running with a speed of 2m/s jumps on a stationary skateboard
of mass 0.5 kg. Will the skateboard move with cat on it if yes find the velocity of board and cat

Answer 1

Proper question:

A cat of body mass 2.5 kg running with a speed of 2 m/s jumps on a stationary skateboard of mass 0.5 kg. Will the skateboard move with cat on it, if yes! Then find the velocity of board and cat.

Provided that:

• Mass of cat = 2.5 kg

• Initial velocity of cat = 2 mps

• Mass of skateboard = 0.5 kg

• Initial velocity of skateboard = 0 mps

To determine:

• Will skateboard move with cat on it?

• The velocity of board and cat!?

Solution:

• Yes, it will move.

• The velocity of board and cat = 5/3 mps

Using concept:

• Law of conservation of momentum

Using formula:

${\small{\underline{\boxed{\sf{\rightarrow \: m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$ denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.)

Required solution:

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 2.5(2) + 0.5(0) = (2.5+0.5)v \\ \\ :\implies \sf 5 + 0 = 3v \\ \\ :\implies \sf 5 \: = 3v \\ \\ :\implies \sf \dfrac{5}{3} \: = v \\ \\ :\implies \sf v \: = \dfrac{5}{3} \: mps$

• Henceforth, their combined velocity is equal to 5/3 mps.

Answer 2

Answer:

Explanation:

Yes, the skateboard will move like a cat with a velocity of 5/3 m/s.

GIVEN: A cat with a body mass of 2.5 kg running with a speed of 2m/s jumps on a stationary skateboard of mass 0.5 kg.

TO FIND: Will the skateboard move with a cat on it if yes then find the velocity of the board and cat system

SOLUTION:

As we are given in the question,

Body mass of Cat = 2.5kg.

Speed of cat = 2m/s

Mass of stationary skateboard =  0.5kg

So now,

m1=2.5kg

u1=2m/s

m2=0.5kg

u2=0

So,

m3=m1+m2=3kg

Now,

According to the law of conservation of linear momentum.

m1u1+m2u2=m3v3

5+0=3v3

v3=5/3m/s

Therefore,

Yes, the skateboard will move like a cat with a velocity of 5/3 m/s.