the arrow is fired is zero because nothing in the system is moving. Therefore, t

Question

the arrow is fired is zero because nothing in the system is moving. Therefore, the total horizontal momentum of the system after the arrow is fired must also be zero. We choose the direction of firing of the arrow as the positive x direction Identifying the archer as particle 1 and the arrow as particle 2, we have m1 = 68.8 kg, m2 = 0.42 kg, and f = 52.9 m/s. Set the final momentum of the system equal to zero: and substitute Solve this equation for numerical values "(68.8 k 329ms 52.9 m/s) The negative sign for indicates that the archer is moving to the left in the figure after the arrow is fired, in the direction opposite the direction of motion of the arrow, in accordance with Newton's third law. Because the archer is much more massive than the arrow, his acceleration and consequent velocity are much smaller than the acceleration and velocity of the arrow MASTER IT HINTS: GETING STARTEDI IMSTUCK What if he misses the target, and pulls out a second arrow (of the same weight as the first) to fire again? This time, he fires it harder. The arrow is measured to have a velocity of 82 m/s by a stationary bystander. This happens while the archer is still sliding backwards at -0.323 m/s. What will be his velocity (in the x-direction) after firing the second arrow? (Notice that his "mass" after firing the second arrow will be 67.96 kg!) v = 1-0501 Determine the relation between the initial and final velocities involved and use it to find the new speed of the archer.

Explanation / Answer

now for second arrow and archer system,

v1i = - 0.323m/s

v2i = - 0.323 m/s

v2f = 82 m/s

Applying momentum conservation,

68.8 x -0.323 = (67.96 v1f) + (0.42 x 82)

v1f = - 0.834 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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