0
$\begingroup$

The question is as follows:

enter image description here

I did the following:
force on m1 = m1g sin(45)
thus acceleration of m1 with respect to m2 = g sin(45)

coming to m2,
force on m2 due to m1 = m1g cos(45)
but this is in diagonal direction, so we need to resolve it.
thus horizontal component = m1gcos(45) * sin(45)
now acceleration of m2 with respect to ground will be force/mass
thus acceleration = m1g cos(45)sin(45) / m2 = g/4 ( on substituting values )

for m1 :
h$√2$ = 1/2at2
again putting values, we have t2 = 2h/5

for m2 : distance = 1/2 * g/4 * 2h/5 = h/2

but correct answer is h/4.
they used concept of center of mass
I prefer my method.
someone please help me understand where I went wrong

$\endgroup$
7
  • $\begingroup$ I get $\frac{h}{3}$ after considering pseudo force. Is the final answer $\frac{h}{4}$ correct? $\endgroup$
    – Iti
    Apr 28, 2021 at 5:44
  • $\begingroup$ @om jogekar I think h/3 is correct. I think you forgot to consider pseudo force. $\endgroup$ Apr 28, 2021 at 5:47
  • 2
    $\begingroup$ We all prefer our original method because we are more comfortable in it. But centre of mass method makes it much much easier to solve this. It will help you for JEE :p $\endgroup$
    – Natru
    Apr 28, 2021 at 6:12
  • $\begingroup$ @Iti yes h/4 is correct. see the solution below $\endgroup$ Apr 28, 2021 at 9:06
  • $\begingroup$ @DheerajKumar h/3 is wrong. see below $\endgroup$ Apr 28, 2021 at 9:06

1 Answer 1

1
$\begingroup$

enter image description here

From the Free Body Diagram of block,
$N_1+m_1a\cos45°=m_1g\cos45°\tag{1}$
(As block moves only on the surface with respect to wedge)
$m_1a\cos45°+m_1g\sin45°=m_1a_1\tag{2}$
From the Free body diagram of wedge,
$N_1\cos45°=m_2a=2m_1a\tag{3}$
From (3),
$N_1=2\sqrt{2}m_1a$

Putting $N_1$ in (1) gives
$a=\frac{g}{5}$

Putting $a$ in (2) gives,
$a_1=\frac{6g}{5\sqrt{2}}$

The block is initially at rest and the hypotenuse of wedge is $\frac{h}{\sin45°}$
So, by $s=ut+\frac{1}{2}a_1t^2$
$\frac{h}{\sin45°}=\frac{1}{2}\frac{6g}{5\sqrt{2}}t^2$
$\implies t^2=\frac{10h}{3g}$

At, the same time $t$ wedge moves by some distance say $s$ leftwards, given by
$s=\frac{1}{2}at^2$
$\implies s=\frac{1}{2}\frac{g}{5}\frac{10h}{3g}$
Hence, $s=\frac{h}{3}$
So distance moved by wedge =$\frac{h}{3}$

$\endgroup$
2
  • $\begingroup$ In the fbd of block, I used the fact that wedge is sttaionary by inserting pseudo force in the equation of block and then it will move only on the surface repative to wedge. $\endgroup$
    – Iti
    Apr 28, 2021 at 14:17
  • 1
    $\begingroup$ i am so sorry for misleading you your answers were right it is h/3 the textbook was wrong my teacher clarified it. $\endgroup$ Apr 28, 2021 at 14:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.