A solid wooden disk of mass 5 kg and radius 10cm sits at the bottom of a 45 degr

Question

A solid wooden disk of mass 5 kg and radius 10cm sits at the bottom of a 45 degree incline. A small 1g projectile traveling horizontally just less than 20 cm above the ground at 12,000 m/s imbeds itself in the top of the disk and comes to rest relative to the wood just below the disk's rim. The friction exerted by the ground on the bottom of the disk is sufficient to keep it from slipping during the collision. As a result of the impact, the wheel now rolls smoothly without slipping up the incline. How far along the surface of the incline does it roll before stopping? Following the collision, you may ignore the mass of the bullet.

Explanation / Answer

angular momentum conservation,

0.001 x 12000 x 0.20 = (0.001 x 0.10^2 + 5 x 0.10^2 /2 ) w

w = 95.96 rad/s

using energy conservation,

Iw^2 + mv^2 /2 = mgh

(0.001 x 0.10^2 + 5 x 0.10^2 /2 )95.96^2 /2 + (5 +0.001) (95.96 x0.10)^2 /2 = (5 + 0.001)x9.8x h

h = 7.05 m

h = lsin45

l = 9.97 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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