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An iron block with mass mB slides down a frictionless hill of height H. At the base of the hill, it collides with and sticks to a magnet with mass mM.What is the speed v of the block and magnet immediately after the collision?

Answer :

Answer:

[tex]v=\sqrt{2gH}\dfrac{mB}{mM+mB}[/tex]

Explanation:

Given that

Mass of first block = mB

Mass of second block = mM

Height = H

The speed of the mass m just before the collision

[tex]V=\sqrt{2gH}[/tex]

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum =  Final linear momentum

Pi = mB V

lets take the final speed of the system becomes v

Pf= (mM+mB) v

m V = (mM+mB) v

[tex]v=\sqrt{2gH}\dfrac{mB}{mM+mB}[/tex]

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