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?

Expert's answer

Let m_{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 energy

m_{1}*V^{2} = (m_{1 }+ m_{2})*g*h

V = 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.

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.

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