Question

Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s respectively.
They collide and after the collision the first object moves at a velocity of 1.67 m./s. Determine the velocity of the second object.


drop some thanks plz I'll return your thanks ​

Answer 1

Answer:

Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s respectively.

They collide and after the collision the first object moves at a velocity of 1.67 m./s. The velocity of second object is 1.165m/s

Explanation:

Momentum of an object is given by,

             $P = mass \times velocity$

According to the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.  

Given,                

Before collision:

• object 1   ,

         mass = 100 g = $\frac{100}{1000} kg = 0.1 kg$   and velocity = 2 m/s                                    

         ( $m_{1} = 0.1 kg$  & $u_{1} = 2 m/s$ )

        momentum of  object 1  = $m_{1} u_{1} = 0.1\times 2 = 0.2 \mathrm{kgms}^{-1}$

• object 2   , mass = 200 g= $\frac{200}{1000} kg = 0.2 kg$ and velocity = 1 m/s                                    

          ($m_{2} = 0.2 kg$  & $u_{2} = 1 m/s$ )

        momentum of  object 2  =  $m_{1} u_{1} = 0.2\times 1 = 0.2 \mathrm{kgms}^{-1}$

Total Momentum before collision =   $m_{1} u_{1}+ m_{2}u_{2}$

                                                        =  $0.2 + 0.2 = 0.4 \mathrm{kgms}^{-1}$

After collision:

• object 1   ,

          mass = 100 g and velocity = 1.67 m/s                                    

          ($m_{1} = 100 g = 0.1 kg$  & $V_{1} = 1.67 m/s$ )

        momentum of  object 1  = $m_{1} V_{1}=0.1 \times 1.67=0.167 \mathrm{kgms}^{-1}$

• object 2   ,

         mass = 200 g and velocity = ?                                    

         ($m_{2} = 200 g = 0.2 kg$  & $V_{2} = ?$ )

          momentum of  object 2  = $m_{2} V_{2} = 0.2\times V_{2} = 0.2 V_{2} \mathrm{kgms}^{-1}$

Total Momentum after collision =   $m_{1} V_{1}+ m_{2}V_{2}$

                                                     $= 0.167+ 0.2 V_{2}$

Using the law of conservation of momentum,

                      $m_{1} u_{1}+m_{2} u_{2} = m_{1} V_{1}+m_{2} V_{2}$

                      $0.4 = 0.167+0.2 V_{2}$

                      $0.2 V_{2} = 0.4-0.167$

                     $V_{2} = \frac{0.4-0.167}{0.2} = 1.165 m/s$

     

Answer 2

Answer:

We can solve the given problem by using the law of conservation of momentum.

Given

Mass of one of the objects (m1) = 100 g = 0.1 kg

Mass of the other object (m2) = 200 g = 0.2 kg

Velocity of m1 before collision (v1) = 2 m/s

Velocity of m2 before collision (v2) = 1 m/s

Velocity of m1 after collision (v3) = 1.67 m/s

Find out

Velocity of m2 after collision= v4

Solution

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

Therefore, m1v1 + m2v2 = m1v3 + m2v4

On sustitututing the known and given values we get

2(0.1) + 1(0.2) = 1.67(0.1) + v4 × (0.2)

0.4 = 0.167 + 0.2 × v4

v4= 1.165 m/s

Answer

Hence, the velocity of the second object becomes 1.165 m/s after the collision

Hope it helps you

Explanation:

Hey

Thanks for thanks ...

btw Are u in class 9th ?

if yes I am too..

Can I ask u Something ... U have answered 1000+ so can you message others users privately ??

Good afternoon :)

(ᵔᴥᵔ) (^_^)