Question

A ball of mass m falls vertically from a height h and collides with a block of equal mass m moving horizontally with a velocity v on a surface.The coefficient of kinetic friction between thr block and the surface is 0.2,while the coefficient of restitution between the ball and block is 0.5.There is no friction acting between the block and the ball.The velocity of block just after collision decreases by

Answer 1

@michele_laino

Answer 2

@ganeshie8

Answer 3

I m not getting how can this be possible
m√2gh=mVy
And from thr coefficient of restitution i got this
e=-Vy/Uy..

Answer 4

I got this first equation by momentum conservation along y axis..

Answer 5

.

Answer 6

I think to have the solution, please tag, so we can start the corresponding discussion

Answer 7

You can get another equation by conservation of linear momentum along x-axis.

Answer 8

@michele_laino

Answer 9

And how can we apply momentum conservation along x axis when we know that there is a force in that direction for the sytem concerned..
(Friction force)

Answer 10

there is an impulsive force between the two objects which takes place during the collision

Answer 11

such force \(F\) does exist for a short time interval \(\Delta t\)

Answer 12

Do u wanna say that
mu-integral of Ndt =mvy

Answer 13

Normla force is providing the necessary impluse to the ball here..

Answer 14

no, please note that
now, the transferred momentum, is:
\[\Delta Q = \frac{{m{v_0}}}{2} = m\sqrt {\frac{{gh}}{2}} \]

Answer 15

since the coefficient of restitution is \(1/2\)

Answer 16

so, I apply the theorem of conservation of kinetic energy, and I can write:
\[\frac{{m{V^2}}}{2} - \mu \left( {mg + F} \right)\Delta tV = \frac{{mV_1^2}}{2}\]

Answer 17

where \(V,\;V_1\) are the velocities of the horizontal moving block before, and after the collision respectively

Answer 18

F??

Answer 19

\(F\) is the impulsive force, which take place during collision, between the two blocks

Answer 20

takes*

Answer 21

next, we can make these approximations:
\[\begin{gathered}
F \gg mg \hfill \\
\hfill \\
V \cong {V_1} \hfill \\
\end{gathered} \]

Answer 22

Sir do u mean to say that it is due to impulse force F that the ball after collision will move in x direction also..??

Answer 23

more precisely \(F\) takes account of the weight of the vertical moving body

Answer 24

yes! Since the collision prevents the horizontally moving body to change is vertical coordinate, which is, and it will be zero

Answer 25

its*

Answer 26

If it takes into account thr vertical motion then how can we apply this .
1/2mv^2- u(mg+F)∆t=1/2mv'^2
U wrote that impulsive force is in vertical direction n u r applying it for horizontal motion of block ..
How??

Answer 27

since the impulsive force causes an increasing of the friction force, which is horizonatlly directed

Answer 28

Sir it should be
Mv-integral of u(mg-F)dt=Mv'

Answer 29

Sorry mg+F

Answer 30

If I neglect \(mg\), then \(F\;\Delta t\) can be considered, with a good approximation, the momentum transferred during collision
now we have the subsequent steps:
\[\begin{gathered}
\frac{m}{2}\left( {{V^2} - V_1^2} \right) = \mu \left( {mg + F} \right)\Delta tV \hfill \\
\hfill \\
\frac{m}{2}\left( {V + {V_1}} \right)\left( {V - {V_1}} \right) = \mu \left( {mg + F} \right)\Delta tV \hfill \\
\hfill \\
\frac{m}{2} \cdot 2V\left( {V - {V_1}} \right) = \mu F\Delta tV \hfill \\
\end{gathered} \]

Answer 31

therefore, I got this velocity change:
\[V - {V_1} = \frac{{\mu F\Delta t}}{m} = \mu \sqrt {\frac{{gh}}{2}} \]

Answer 32

I got v-v1 as mu√gh/2

Answer 33

it is right then!

Answer 34

That gives us the ans as 0.1√2gh

Answer 35

Correct??

Answer 36

yes! If we rationalize, we get:
\[V - {V_1} = \frac{{\mu F\Delta t}}{m} = \mu \sqrt {\frac{{gh}}{2}} = \frac{{0.2\sqrt 2 }}{2}\sqrt {gh} = 0.1\sqrt 2 \sqrt {gh} \]

Answer 37

Bt ans is diff in ans book..

Answer 38

They say it's 0.3√2gh

Answer 39

Gudnyt sir!! Gotto go now!

Answer 40

ok! goodnight!

Answer 41

If you get the ans they gave then pls do inform me