Question

A bullet of mass 20g moving with a speed of 100m/s hits a stationary wood block of mass 2kg free to move and remains in the block find the speed of the block containing a bullet

Answer 1

The answer is 0.99 m/s

Explanation:

m1 = 20 g

m2 = 2kg + 20g =2020 g

v1 = 100m/s

v2 = ????

we know,

m1v1=m2v2

or, v2 =m1v1/m2

or, v2 = 20*100 /2020

or, v2 = 0.99

Answer 2

The final speed of the block containing a bullet is 0.99 m/s.

Explanation:

It is given that,

Mass of the bullet, m = 20 g = 0.02 kg

Speed of the bullet, u = 100 m/s

Mass of the wood, m' = 2 kg

The wood is at rest, u' = 0

Let V is the speed of the block containing a bullet. As the bullet remains in the block. The value of V can be calculated using the conservation of momentum as :

$mu+m'u'=(m+m')V$

$mu=(m+m')V$

$V=\dfrac{mu}{(m+m')}$

$V=\dfrac{0.02\times 100}{(0.02+2)}$

V = 0.99 m/s

So, the final speed of the block containing a bullet is 0.99 m/s. Hence, this is the required solution.

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Conservation of momentum

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