Question

A particle of mass 10g is moving with a velocity 2i+j+3ť (cm/s) magnitude of its linear momentum is​

Answer 1

Given,

Mass of a particle=10g

Its velocity=2i^+j^+3t^ cm/s

To Find,

Its linear momentum.

Solution,

Mass=10g

Velocity=2i^+j^+3t^cm/s

So, linear momentum of particle is

P=mass× velocity

So, P = 10 ×(2i^ + j^+ 3k^)

         =20i^+10j^+30k^

hence this is linear momentum vector=20i^+10j^+30k^

Now Magnitude of linear momentum

P =√20²+10²+30²

  =√400+100+900

  =√1400g cm/s

P = √1400 g cm/s

Hence, magnitude of linear momentum is √(1400) g cm/s.

Answer 2

Answer:

Particle of mass 10 g moving with velocity (2i+j+3k) has linear momentum of magnitude √(1400) g cm/s.

Explanation:

linear momentum of particle is  

   P= MV  

where,   M= mass of object        

       V = velocity

Given : M = 10 g     V = 2i + j+ 3k  cm/s

so,    

 P = 10 (2i + j+ 3k)  

  P = (20i + 10j + 30k) g cm/s  

hence this is linear momentum vector

Now Magnitude of linear momentum

 P =$\sqrt{20^{2}+10^{2} + 30^{2} }$  

P = $\sqrt{1400}$ g cm/s

hence magnitude of linear momentum is √(1400) g cm/s.