Question

What is the momentum of an object of mass 300 g moving with a velocity of 18 km/h ? (1) 5 kg m/s 1.5 kg m/s 2.5 kg m/s 0.5 kg m/s.​

Answer 1

Answer:

Initial velocity (u)=

Initial velocity (u)= 5

Initial velocity (u)= 520

Initial velocity (u)= 520

Initial velocity (u)= 520 =4m/s

Initial velocity (u)= 520 =4m/sAcceleration produced =

Initial velocity (u)= 520 =4m/sAcceleration produced = m

Initial velocity (u)= 520 =4m/sAcceleration produced = mF

Initial velocity (u)= 520 =4m/sAcceleration produced = mF

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E =

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 2

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u 2

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u 2 )=

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u 2 )= 2

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u 2 )= 21

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u 2 )= 21

Initial velocity (u)= 520 =4m/sAcceleration produced = mF =1m/s 2 Final velocity(v)=u+at=4+1(10)=14So, Change in K.E = 21 m(v 2 −u 2 )= 21 (5)(196−16)=450J

Answer 2

Answer:

Given :-

• An object of mass 300 g moving with a velocity of 18 km/h.

To Find :-

• What is the momentum of an object.

Formula Used :-

$\clubsuit$ Momentum Formula :

$\mapsto \sf\boxed{\bold{\pink{Momentum =\: Mass \times Velocity}}}$

Solution :-

First, we have to convert mass g to kg :

$\implies \sf Mass =\: 300\: g$

$\implies \sf Mass =\: \dfrac{3\cancel{00}}{10\cancel{00}}\: kg$

$\implies \sf Mass =\: \dfrac{3}{10}\: kg$

$\implies \sf\bold{\purple{Mass =\: 0.3\: kg}}$

Now, again we have to convert velocity :

$\implies \sf Velocity =\: 18\: km/h$

$\implies \sf Velocity =\: {\cancel{18}} \times \dfrac{5}{\cancel{18}}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf\bold{\purple{Velocity =\: 5\: m/s}}$

Now, we have to find the momentum of an object :

Given :

• Mass = 0.3 kg
• Velocity = 5 m/s

According to the question by using the formula we get,

$\longrightarrow \sf Momentum =\: (0.3)(5)$

$\longrightarrow \sf Momentum =\: 0.3 \times 5$

$\longrightarrow \sf\bold{\red{Momentum =\: 1.5\: kg\: m/s}}$

${\small{\bold{\underline{\therefore\: The\: momentum\: of\: an\: object\: is\: 1.5\: kg\: m/s\: .}}}}$

Hence, the correct options is option no (2) 1.5 kg m/s.