You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Com in Perfectly Inelastic Collision

A bullet of m...

Question

A bullet of mass 300grams moving with velocity $v$ strikes a wooden block of mass 2.7 kg and gets embedded in it. If the system travels the distance of 20 m on the surface and comes to rest, the velocity $v$ is (coefficient of kinetic friction $= 0.25$ and $g = 10 m s^{- 2}$ ):

200 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

300 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 m s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

50 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $100 m s^{- 1}$
From conservation of momentum, $(2.7 + 0.3) v_{c o m m o n} = 0.3 v$
$\Rightarrow v_{c o m m o n} = \frac{v}{10}$
$\frac{1}{2} m v_{c o m m o n}^{2} = F . s$
as, $0.5 \times (2.7 + 0.3) \times v_{c o m m o n}^{2}) = μ \times (2.7 + 0.3) g \times 20$
$\Rightarrow v_{c o m m o n} = 10$
From the first equation, $v = 100 m / s$

Suggest Corrections

Similar questions

Q. A bullet of mass

1 \times 10^{- 2}

k g

moving horizontally with a velocity of

2 \times 10^{2} m s^{- 1}

strikes a block of mass

1.99 k g

and gets embedded into it. If the coefficient of kinetic friction between the block and the horizontal surface is

0.25

, then the distance moved by the block with the bullet before coming to rest is

A bullet of mass 20 grams moving with a velocity of $100 m s^{- 1}$ strikes a wooden block of mass 80 grams and gets embedded into it. What is the velocity acquired by the block?

Q. A bullet of mass

50 g

moving horizontally with a velocity

210 {m s}^{- 1}

gets embedded in a block of mass

1 k g

kept on a rough horizontal surface. If the coefficient of kinetic friction between the block and the surface is

0.5

, the block-bullet system will move a distance of ____________ before coming to rest. (Acceleration due to gravity

= 10 {m s}^{- 2}

)

A bullet of mass $20 g$ moving with a velocity of $100 m s^{- 1}$ strikes a wooden block of mass $80 g$ at rest and gets embedded into it. Calculate the velocity of the combined system.

Q. A bullet of mass 0.01 kg is fired horizontally into a 4kg wooden block at rest on horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.25. The bullet gets embedded in the block and the combination moves 20 m before coming to rest. With what speed did the bullet strike the block?

Join BYJU'S Learning Program

A bullet of mass 300grams moving with velocity v strikes a wooden block of mass 2.7 kg and gets embedded in it. If the system travels the distance of 20 m on the surface and comes to rest, the velocity v is (coefficient of kinetic friction=0.25 and g=10ms−2):

The correct option is D 100 ms−1From conservation of momentum, (2.7+0.3)vcommon=0.3v⇒vcommon=v1012mv2common=F.sas, 0.5×(2.7+0.3)×v2common)=μ×(2.7+0.3)g×20⇒vcommon=10From the first equation, v=100m/s

A bullet of mass 300grams moving with velocity $v$ strikes a wooden block of mass 2.7 kg and gets embedded in it. If the system travels the distance of 20 m on the surface and comes to rest, the velocity $v$ is (coefficient of kinetic friction $= 0.25$ and $g = 10 m s^{- 2}$ ):