Question

Two balls a and b of masses 100g and 250g connected to the ends of a weightless stretched spring are placed on a smooth horizontal surface .on being released the ball b goes towards west with an initial acceleration of 10cm/s^2. with what initial acceleration and in which direction will the ball a go?

Answer 1

For block B we can say

$F_b = m_b * a$

given that

$m_b = 250g = 0.250 kg$

$a = 10 m/s^2$

now by above formula

$F = ma = 0.250 * 10 = 2.50 N$

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

$F = ma$

$2.5 = m_a * a$

given that

$m_a = 100g = 0.100 kg$

now by above equation

$2.5 = 0.100 * a$

solving above equation

$a = 25 m/s^2$

so acceleration of block "a" is 25 m/s^2 towards EAST

Answer 2

Answer:

Explanation:

For block B we can say

given that

now by above formula

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

given that

now by above equation

solving above equation

so acceleration of block "a" is 25 m/s^2 towards EAST

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