“Laura, whose mass is 35 kg, jumps horizontally off a 55 kg canoe at 1.5 m/s relative to the canoe. What is the canoe’s speed just after she jumps?”
I’m trying to use conservation of momentum to solve this, so m1v1 = m2v2, or (35)(1.5) = (55)(vf_canoe).
Solving for the velocity of the canoe, I get 0.955 m/s, but apparently the answer is incorrect even when you stick a negative sign in front of it. I really don’t know how else to approach this problem, though I did assume that both Laura and the canoe are initially at rest. However, maybe “relative” to the canoe has some significance? If the canoe was initially moving, the problem sure didn’t provide any specific details. I’m really stuck…
4 Answers
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You have missed the key.
Laura jumps at 1.5 m/s RELATIVE TO THE CANOE.
( not relative to the centre of mass which also happens to coincide with the earth.)
Hence we can calculate relative to the centre of mass instead.
35 V1 = 55 V2 where V1 and V2 are the magnitudes of the two speeds. (momentum is conserved)
and V1 + V2 = 1.5
hence V2= 1.5 -V1
Now substitute back into momentum
35 V1 = 55( 1.5-V1)
= 55 * 1.5 – 55V1
35 V1 + 55 V1 = 82.5
90V1 = 82.5
V1= 0.917 m/s
V2= 1.5 – V1 = 0.583 m/s
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Here, the speed is measured relative to the canoe and not relative to the ground.
The observer is in the canoe itself. He measures the velocity of Laura. The velocity of canoe
relative to itself is zero.
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since the speed is relative to coneo,
so let speed of laura be X,
then speed of canoe is (1.5-X)
now apply conservation of momentum,
35X-55(1.5-X) = 0,
solving X= 0.9167m/s
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relative or not,this is the only speed you have…
so,after jumping,the canoe might be 55-35=20kg,`cos laura is away…
it`s not said exactly…
so
55×1.5=20xX
X=4,125
