From the top of a tower of the height $100 m$ , a $10 g m$ block is dropped freely, and a $6 g m$ bullet is fired vertically upwards from the foot of the tower with velocity $100 m s^{- 1}$ simultaneously. They collide and stick together. The common velocity after collision is $(g = 10 m s^{- 2})$

27.5 m s^{- 1}

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

150 m s^{- 1}

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

40 m s^{- 1}

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

100 m s^{- 1}

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

Open in App

Solution

The correct option is B $27.5 m s^{- 1}$
Their relative velocity is $v_{r} = 100 m / s$ and relative acceleration $a_{r} = 0$
So they collide at $t$ seconds later then
$s$ = $v_{r} t + \frac{1}{2} a_{r} t^{2}$
$100 = 100 t + 0$
$t = 1 s e c$
After 1 sec
$V_{b l o c k} = g t$
$V_{b l o c k} = 10$
$V_{b u l l e t} = 100 - g t$
$V_{b u l l e t} = 90$
Using conservation of momentum
$m_{b l o c k} V_{b l o c k} + m_{b u l l e t} V_{b u l l e t} = m_{c o m b i n e d} V_{c o m b i n e d}$
$.01 x 10 - .006 x 90 = .016 x V_{c o m b i n e d}$
$V_{c o m b i n e d} = - 27.5$
Hence common velocity after collision will be $27.5 m / s$ upwards.

Suggest Corrections

Similar questions

Q. From the top of tower of height 100 m a 10 gm block is dropped freely and a 6 gm bullet is fired vertically upward from the foot of the tower with velocity 100

m s^{- 1}

simultaneously. They collide and stick together after time. Find the time after which they collide with each other.

Q. A wooden block of mass

10 g

is dropped from the top of a tower

100 m

high. Simultaneously, a bullet of mass

10 g

is fired from the foot of the tower vertically upwards with a velocity of

100 m s^{- 1}

. If the bullet is embedded in it, how high will it rise above the tower before it starts falling?

(

Consider

g = 10 m s^{- 2})

Q. A wooden block of mass 10 gm is dropped from the top of a cliff 100 m high. Simultaneously a bullet of same mass is fired from the foot of the cliff vertically upwards with a velocity of

100 m s^{- 1}

. If the bullet after collision gets embedded in the block, the common velocity of the bullet and the block immediately after collision is

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

Q. A wooden block is dropped from a tower of height 200 m. Simultaneously a bullet is fired from the foot of the tower aiming the block with the velocity of

100 {m s}^{- 1}

. Given that the masses of both the block and the bullet , each equal to 100 g and the bullet is embedded in the block, find the maximum potential energyof the system of the block and the bullet. (

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

) (Neglect the time of impact).

A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff upward with a velocity of $100 m s^{- 1}$ . At what time will the bullet and the block meet ? Take g= $9.8 m s^{- 2}$ .

Join BYJU'S Learning Program

From the top of a tower of the height 100m, a 10gm block is dropped freely, and a 6gm bullet is fired vertically upwards from the foot of the tower with velocity 100ms−1 simultaneously. They collide and stick together. The common velocity after collision is (g=10ms−2)

From the top of a tower of the height $100 m$ , a $10 g m$ block is dropped freely, and a $6 g m$ bullet is fired vertically upwards from the foot of the tower with velocity $100 m s^{- 1}$ simultaneously. They collide and stick together. The common velocity after collision is $(g = 10 m s^{- 2})$