Question

Two balls $A$ and $B$ of masses $100gm$ and $250gm$ respectively are connected by a stretched spring of negligible mass and placed on a smooth table. When the balls are released simultaneously the initial acceleration of $B$ is $10cm/{sec}^2$ west ward. What is the magnitude and direction of initial acceleration of the ball $A$.

• A. $25cm/{sec}^2$ East ward
• B. $25cm/{sec}^2$ North ward
• C. $25cm/{sec}^2$ West ward
• D. $25cm/{sec}^2$ South ward

Answer 1

The correct option is A $25cm/{sec}^2$ East ward

For block B we can say

$F_b=m_b\times a$

given that

$m_b=250g=0.250kg$

$a=10m/s^2$

Now by above formula

$F=ma=0.250\times 10=2.50N$

Now we know that both blocks are connected by same spring

So spring force will be same on both blocks

so force block $"A"$ the spring force will be same

$F=ma$

$2.5=m_a\times a$

given that

$m_a=100g=0.100kg$

now by above equation

$2.5=0.100\times a$

solving above equation

$a=25m/s^2$

so$,$ acceleration of block $"a"$ is $25m/s^2$ towards East

Hence,

option $\left(A\right)$ is correct answer.