Open-Response Homework Problem 10.5 An archer shoots an arrow with an initial ve

ID: 586021 • Letter: O

Question

Open-Response Homework Problem 10.5 An archer shoots an arrow with an initial velovity of 55m/s at a thin

target which is initially moving toward him at 5m/s . The arrow then strikes and passes through the target and

continues travelling. The mass of the target is 0.6 kg, the mass of the arrow is 0.3 kg, and the coefficient of

restitution for this collision is e = 0.75 Yes, it is negative since the arrow passes through the target. DON’T

PANIC. All of your equations and concepts still hold true for this type of collision. Use the equation for coefficient

of restitution that is on your equation sheet. Pay very careful attention to the signs on your velocities.

(a)[11 pts] Find the velocities of the arrow and the target after the collision.

(b)[4 pts] Before the collision, how much energy is convertible to other forms?

(c)[5 pts] After the collision, what percent of the convertible kinetic energy was converted?

ma = mass of arrow = 0.3 kg

Vai = initial velocity of arrow = 55 m/s

mt = mass of the target = 0.6 kg

Vti = initial velocity of the target = - 5 m/s

Vaf = final velocity of the arrow

Vtf = final velocity of the target

Using conservation of momentum

ma Vai + mt Vti = ma Vaf + mt Vtf

0.3 (55) + (0.6) (-5) = 0.3Vaf + 0.6Vtf

0.3Vaf + 0.6Vtf = 13.5

Vaf + 2 Vtf = 45 eq-1

Using the formula

e = Vtf - Vaf / (Vai - Vti)

- 0.75 = Vtf - Vaf / (55 - (-5))

Vtf - Vaf = -45

Vtf = Vaf - 45 eq-2

Using eq-1 and eq-2

Vaf + 2 (Vaf - 45 )= 45

Vaf = 45 m/s

Vtf = 0 m/s

Total energy before collision = (0.5)ma V2ai + (0.5) mt V2ti = (0.5) (0.3 (55)2 + (0.6) (-5)2) = 461.25 J

Total energy aftre collision = (0.5)ma V2af + (0.5) mt V2tf = (0.5) (0.3 (45)2 + (0.6) (0)2) = 303.75 J

percentage energy converted = 303.75 x 100 /461.25 = 65.85 %

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