You fire a 12 g bullet horizontally into a 15 kg brick which itself is connected

Question

You fire a 12 g bullet horizontally into a 15 kg brick which itself is connected to a horizontal spring. The bullet has an initial speed of 340 m/s and the spring has a spring constant of 650 N/m. (a) Use the conservation of momentum to find the speed of the bullet/brick after they collide. Assume the brick is at rest initially and the bullet fully embeds in the brick upon contact. (b) Use energy conservation to find the distance that the spring is compressed. Further assume that the counter on which the brick sits is frictionless.

Explanation / Answer

Given

mass of bullet mb = 12 g = 0.012 kg,

mass of brick mB = 15 kg

speed of bullet is vb = 340 m/s

spring constant k = 650 N/m

here considering the conservation of momentum of bullet and brick system

that is before collision and after collision the momentu is same

after the collision the bullet embeded in the brick, both move with common speed V so

a) CONSERVATION OF MOMENTUM

mb*vb+mB*vB = (mb+mB)V

V = (mb*vb+mB*vB )/(mb+mB)

V = (0.012*340+15*0)/(0.012+15) m/s

v = 0.27178 m/s

both will move with 0.27178 m/s and collide with the spring so that the spring will compress

b) CONSERVATION OF ENERGY

The kinetic energy of the bullet+ brick system will impart to the spring so that the spring will compress,resulting the elastic potential energy will develop in the spring .

Equating these energies

0.5*(mb+mB)V^2 = 0.5*K*x^2

where x is compression in the spring

x^2 = (mb+mB)V^2 / K

x = sqrt((mb+mB)V^2 / K)

x = sqrt(0.012+15)(0.27178)^2/650) m

x = 0.0004402921169 m

x = 0.4403 mm

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

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