Question

Two small magnets are placed at points
and B on a frictionless horizontal
table Their masses are 20 8 and 12
respectively. The magnets are released
and they move on the table due to mutual
attraction. When the first magnet reaches
a point X, its speed is found to be 3 m/s
Find the speed of the other magnet at
that instant​

Answer 1

The the speed of the other magnet at  that instant​ is 4.74 m/s.

Explanation:

Given that,

Mass of first magnet = 20 kg

Mass of second  magnet = 8 kg

Distance = 12 m

Speed of first magnet= 3 m/s

We need to calculate the speed of the other magnet at  that instant​

Using conservation of energy

$P.E=K.E$

$\dfrac{GmM}{r^2}=\dfrac{1}{2}mv_{1}^2+\dfrac{1}{2}Mv_{2}^2$

Put the value into the formula

$\dfrac{6.67\times10^{-11}\times20\times8}{12^2}=\dfrac{1}{2}\times20\times9+\dfrac{1}{2}\times8\times v^2$

$v^2=-22.475\ m/s$

$v=-4.74\ m/s$

Negative sign shows the opposite direction of motion.

Hence, The the speed of the other magnet at  that instant​ is 4.74 m/s.

Learn more :

Topic : conservation energy

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