# Answer to Question #3848 in Mechanics | Relativity for mia

Question #3848

A 0.73 kg magnetic target is suspended on a string. A 0.025 kg magnetic dart, shot

horizontally, strikes it head on. The dart and the target together swing up 12 cm above the

initial level. What was the initial velocity of the dart?

horizontally, strikes it head on. The dart and the target together swing up 12 cm above the

initial level. What was the initial velocity of the dart?

Expert's answer

Let m

V - initial velocity of the dart.

Consider the target and the dart didn't have the initial potential energy,

let's use the law of conservation of energy:

E

where

E

E

E

E

m

V = sqrt((m

V = sqrt((0.025+0.73)*9.8*0.12/0.025) = 5.96 m/s.

So, the initial velocity of the dart is 5.96 m/s.

_{1}be the mass of the magnetic dart, m_{2}- mass of the target.V - initial velocity of the dart.

Consider the target and the dart didn't have the initial potential energy,

let's use the law of conservation of energy:

E

_{1}= E_{2},where

E

_{1}- initial energy of the system,E

_{2}- energy of the system after the collision.E

_{1}= E_{k}- kinetic energy,E

_{2}= E_{p}- potential energym

_{1}*V^{2}= (m_{1 }+ m_{2})*g*hV = sqrt((m

_{1}+m_{2})*g*h/m_{1})V = sqrt((0.025+0.73)*9.8*0.12/0.025) = 5.96 m/s.

So, the initial velocity of the dart is 5.96 m/s.

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