Question

A cat lunges towards its unsuspecting owner. Treat the cat as a point mass of mass $m=2kg$ and the owner as a stationary point mass of mass $M=60kg$ located at $y=0m$.
If the cat has velocity $v_0=0m/s$ at height $y_0=2.0m$ at $t=0sec$, when will the cat-owner system have a center-of-mass height of $y_{CM}=0.01m$?

• A. Never
• B. $t=0.0sec$
• C. $t=0.42sec$
• D. $t=0.59sec$
• E. $t=0.64sec$

Answer 1

Answer: (D)

$t=0.59sec$

Let the height of the cat at the moment be $h$.

Then the center of mass is given by

$\frac{(2\times h)+(60\times 0)}{60+2}=0.01$

$⟹h=0.31m$

Hence the distance travelled by the cat=$2-0.31=1.69m$

Hence from equation of motion,

$x=\frac{1}{2}g{t}^2$
$⟹t=0.59s$