Question

24. A bullet of mass 'm,' moving with a speed 'x' strikes a wooden block of mass 'm' and gets embedded in it. The speed of the embedded block will be ​

Answer 1

Answer:

xm'/(m' + m)

Step-by-step explanation:

Using the conservation of momentum:

Initially:   mass = m'  and  speed = x

               momentum = m' *x

                                    = xm'

FInally:   mass = mass of bullet + block

                        = m' + m

Let the speed of the block be 'y'

              momentum = (m' + m)y

Initial momentum = final momentum

⇒ xm' = (m' + m)y

⇒ xm'/(m' + m) = y

Hence, the speed of the embedded block will be xm'/(m' + m).

Answer 2

Given :-

A bullet of mass 'm,' moving with a speed 'x' strikes a wooden block of mass 'm' and gets embedded in it

To Find :-

The speed of the embedded block will be ​

Solution :-

Let the masses be m1 and m2

Now

m1u1 + m2u2 = m1v1 + m2v2

We have

m1 = m1

u1 = x

m2 = m2

u2 = 0

m1 = m1 = m2

(m1)(x) + (m2)(0) = (m1 + m2)v

m1x + m2 = m1x + m2x

m2 = m1x - m1x + m2x

m2 = m2x

Hence

Speed will be m2 × x