A bullet of mass 10 g travels horizontally with speed of $100 m s^{- 1}$ and is absorbed by a wooden block of mass 990 g suspended by a string. Find the vertical height through which the block rises, $t a k e g = 10 m s^{- 2}$ .

Open in App

Solution

Momentum before and after striking block-

$10 g \times 100 = (990 + 10) g \times v$ where v is velocity of bullet and block after bullet gets embedded

$⟹ v = 1 m / s$

And final mass is $m = 990 + 10 = 1000 g = 1 k g$

Now, we know from conservation of energy-

$\frac{1}{2} m v^{2} = m g h$

$⟹ h = \frac{v^{2}}{2 g} = \frac{1}{2 \times 10}$

$⟹ h = 0.05 m = 5 c m$

Suggest Corrections

Similar questions

Q. A block of wood of mass

9.8 k g

is suspended by a string. A bullet of mass

200 g m

strikes horizontally with a velocity of

100 m s^{- 1}

and gets imbedded in it. To what height will the block rise?

(g = 10 m s^{- 2})

Q. A bullet fired horizontally with a speed of

400 m / s e c

. It strikes a wooden block of mass

2 k g

hanging vertically with the help of long string. After striking with bullet, block rises a height of

20 c m

. If speed with which bullet emerges out from block is

10 λ

then find

λ

. Mass of bullet is

20 g m

. (take

g = 10 m / s^{2}

)

A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string.The bullet crosses the block and emerges on the other side.If the centre of mass of the block rises through a height of 20.0 cm,find the speed of the bullet as it emerges from the block.

Q. A bullet of mass 10g moving horizontally with a velocity of 400 m/s strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity the block is found to rise a vertical distance of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be:

Q. A bullet of mass 10g moving horizontally with a velocity of 400 ms

^{- 1}

strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be :

Join BYJU'S Learning Program

A bullet of mass 10 g travels horizontally with speed of 100ms−1 and is absorbed by a wooden block of mass 990 g suspended by a string. Find the vertical height through which the block rises, take g=10ms−2.

Momentum before and after striking block-10g×100=(990+10)g×v where v is velocity of bullet and block after bullet gets embedded⟹v=1m/sAnd final mass is m=990+10=1000g=1kgNow, we know from conservation of energy-12mv2=mgh⟹h=v22g=12×10⟹h=0.05m=5cm

A bullet of mass 10 g travels horizontally with speed of $100 m s^{- 1}$ and is absorbed by a wooden block of mass 990 g suspended by a string. Find the vertical height through which the block rises, $t a k e g = 10 m s^{- 2}$ .