Blocks A and B are resting on a smooth horizontal surface given equal speeds of 2ms- 1 in opposite sense as shown in the figure. At t=0 the position of blocks are shown, then the co-ordinates of center of mass at t=3s will be :-

Question

Blocks A and B are resting on a smooth horizontal surface given equal speeds of 2ms- 1 in opposite sense as shown in the figure. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-vpmzmmsmnwbpr6gz.jpg" /> At t=0 the position of blocks are shown, then the co-ordinates of center of mass at t=3s will be :- (A) (1 , 0) (B) (3 , 0) (C) (5 , 0) (D) (2 . 25 , 0)

Tardigrade Admin · Accepted Answer

xcmf=xcmi+vc m× t Here vc m=(4 × 2 - 4 × 2/4 + 4)=0 xC M=(4 × 0 + 4 × 4 . 5/4 + 4)=2.25m yc m=0

Q. Blocks $A$ and $B$ are resting on a smooth horizontal surface given equal speeds of $2 m s^{- 1}$ in opposite sense as shown in the figure.

At $t = 0$ the position of blocks are shown, then the co-ordinates of center of mass at $t = 3 s$ will be :-

$(1, 0)$

$(3, 0)$

$(5, 0)$

$(2.25, 0)$

$x_{c m_{f}} = x_{c m_{i}} + v_{c m} \times t$
Here $v_{c m} = \frac{4 \times 2 - 4 \times 2}{4 + 4} = 0$
$x_{CM} = \frac{4 \times 0 + 4 \times 4.5}{4 + 4} = 2.25 m$
$y_{c m} = 0$

Q. Blocks A and B are resting on a smooth horizontal surface given equal speeds of 2ms−1 in opposite sense as shown in the figure. At t=0 the position of blocks are shown, then the co-ordinates of center of mass at t=3s will be :-

(1,0)

(3,0)

(5,0)

(2.25,0)