A system consists of a 4.0-kg cart and a 1.2-kg cart attached to each other by a

Question

A system consists of a 4.0-kg cart and a 1.2-kg cart attached to each other by a compressed spring. Initially, the system is at rest on a low-friction track. When the spring is released, an explosive separation occurs at the expense of the internal energy of the compressed spring. The change in the spring's internal energy during the separation is 1.0 kJ. A- What is the speed of 1.2-kg cart right after the separation? Express your answer with the appropriate units. B- What is the speed of 4.0-kg cart right after the separation? Express your answer with the appropriate units.

Explanation / Answer

m1 = 4 kg
m2 = 1.2 kg

here,

from conservation of Energy we have :

1000 = 0.5*4*v1'^2 + 0.5*1.2*v2'^2
v2'^2 = (1000 - 2*v1'^2)/0.6 -------------(1)

also from conservation of momentum

m1v1' = m2v2'
v2' = (4/1.2)*v1'
v2' = -3.33*v1'

therefore Eqn 1 becomes :

0.6(-3.33*V1')^2 = 1000 - 2*v1'^2
6.653*v1'^2 + 2*v1'^2 = 1000
v1' = -14.66 m/s

so,

v2'^2 = (1000 - 2*(-14.66)^2)/0.6
v2' = 30.826 m/s

A)
speed of cart 2 after collision = 30.826 m/s

B)
speed of cart 1 after collisin is -14.66 m/s where - sign show direction of cart 1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.