The figure shows two 22.7 kg ice luges that are placed a short distance apart, o

Question

The figure shows two 22.7 kg ice luges that are placed a short distance apart, one directly behind the other. A 3.63 kg tomcat initially standing on one luge jumps to the other one and then back to the first. Both jumps are made as a speed of 3.05 m/srelative to the ice.

What are the final speeds of (a) the first luge and (b) the other luge?

where vs12 is the final velocity of the 1st-luge.

The figure shows two 22.7 kg ice luges that are placed a short distance apart, one directly behind the other. A 3.63 kg tomcat initially standing on one luge jumps to the other one and then back to the first. Both jumps are made as a speed of 3.05 m/srelative to the ice. What are the final speeds of (a) the first luge and (b) the other luge? where vs12 is the final velocity of the 1st-luge.

Explanation / Answer

for first jump he leaves and first luge will gain speed to the left

0 = 3.63*3.05 + 22.7*v

v = -0.488 m/s

then he will land o nthe second it will gain a speed

3.63*3.05 = (3.63+22.7)*v

v=0.42 m/s

then we jumps off the second luge

(3.63+22.7)*0.42 = 3.63*(0.42-3.05) + 22.7*v

v second luge = 0.9077 m/s (this is answer for b)

then he lands on first luge

22.7*-0.488 + 3.63*(0.42-3.05) = (22.7+3.63)*v

v=-0.7833 m/s this is answer for a

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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