) The figure shows two 22.7 kg ice luges that are at rest and placed a short dis

Question

) The figure shows two 22.7 kg ice luges that are at rest and 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 at a speed of 3.05 m/s

relative to the ice. What are the final speeds of (a) the first luge and (b) the second luge?

The figure shows two 22.7 kg ice luges that are at rest and 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 at a speed of 3.05 m/s relative to the ice. What are the final speeds of (a) the first luge and (b) the second 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

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