Question

149. A cube of mass m and height H slides with a speed v. It
strikes the obstacle of height h=--The speed of the
CM of the cube just after the collision is
​

Answer 1

Given that,

Mass of cube = m

height = H

Speed = v

height of obstacle = h

Initially the cube is moving in a speed of v.

The center of mass is at height$\dfrac{H}{2}$.

The obstacle has height  $h=\dfrac{H}{4}$.

The moment arm to the top of the obstacle is

$r=\dfrac{H}{2}-\dfrac{H}{4}$

We need to calculate the initial angular momentum to the top of the obstacle

Using formula of angular momentum

$L_{i}=mvr$

Put the value of r

$L_{i}=mv\dfrac{H}{4}$....(I)

After strikes the obstacle, the cube would rotate about the top of the obstacle.

We need to calculate the final angular momentum

Using formula of angular momentum

$L_{f}=I\omega$

$L_{f}=(I_{cm}+Id^2)\omega$

Put the value of moment of inertia

$L_{f}=(\dfrac{mH^2}{6}+md^2)\omega$...(II)

Where, d = distance from the center of mass of the cube to the top of the obstacle

We need to calculate the value of d

Using pythagorean theorem

$d=\sqrt{(\dfrac{H}{2})^2+(\dfrac{H}{2}-h)^2}$

Put the value of h

$d=\sqrt{(\dfrac{H}{2})^2+(\dfrac{H}{2}-\dfrac{H}{4})^2}$

$d=\sqrt{\dfrac{H^2}{4}+\dfrac{H^2}{16}}$

$d=\sqrt{\dfrac{5H^2}{16}}$

$d=\dfrac{\sqrt{5}H}{4}$

We need to calculate the angular velocity

Using conservation of momentum

$L_{i}=L_{f}$

$mvr=I\omega$

Put the value into the formula

$mv\dfrac{H}{4}=(\dfrac{mH^2}{6}+md^2)\omega$

$v\dfrac{H}{4}=(\dfrac{H^2}{6}+\dfrac{5H^2}{16})\omega$

$\dfrac{v}{4}=\dfrac{23H}{48}\times\omega$

$\omega=\dfrac{12v}{23H}$

We need to calculate the speed of the center of mass

Using formula of velocity

$v' = d\ \omega$

Put the value into the formula

$v'=\dfrac{\sqrt{5}H}{4}\times\dfrac{12v}{23H}$

$v'=\dfrac{3\sqrt{5}}{23}v$

Hence, The speed of the center of mass is $\dfrac{3\sqrt{5}}{23}v$

Answer 2

v1 = √56. v

Explanation:

mv (H/2 - h) = m [ H² + H² / 12 = (H/2)² + (H/2 - h)²] ω

ω = (H/2 - h) v / H²/6 + H²/4 + (H/2 - h)²

v = rω = (H/2 - h)v. √H²/4 + (H/2 - h)²

If h = H/4, then:

v1 = Hv/4 . √H²/4 + H²/16 / (5/12 + 1/16)H²

v/4.√5/4 / (15+3 / 48) = 3√5 v / 18

v1 = √56. v