Question

A block of mass 1 kg is pushed on a movable wedge of mass 2kg and height 30cm with a velocity 6ms

Answer 1

what is to be found??

Answer 2

Answer:

Image

Velocity of block at c,

${ v }^{ 2 }={ u }^{ 2 }+2as$

${ v }^{ 2 }\quad =\quad 6-2\mu g{ x }^{ 2 }$

${ v }^{ 2 }\quad =\quad 36-40\mu$

When the body reaches height h its velocity becomes zero

Therefore K.E= P.E

$P.E\quad =\quad mgh\quad =\quad 1\times 10\times 0.3\quad =\quad 3J$

By conservation of momentum, $m_1\mu_1 +m_2\mu_2= m_2v_2(initial)+ m_2v_2(final)$

$1\times v+m_2(0)\quad =\quad m_1(0)+2v_2$

$v_2\quad =\quad \frac { v }{ 2 }$

here $v_2$ is the velocity of wedge and v is the velocity of the block

By taking square on both sides, we get

${ (v_2) }^{ 2 }\quad =\quad \frac { v_2 }{ 4 }$

K.E of block got converted in to P.E of block+ K.E of wedge

$\frac { 1 }{ 2 } \times 1\times (36-40\mu)\quad =\quad 3(P.E)+\frac { 1 }{ 2 } \times 2\times 36-\frac { 40\mu }{ 4 }$

by simplifying we get,,

$36-40\frac { \mu }{ 2 } \quad =\quad 6$

$\mu\quad =\quad \frac { 18 }{ 40 } \quad =\quad 0.45$