Asked by help
A 45.00 kg child standing on a frozen pond throws a 0.60 kg stone to the east with a speed of 5.80 m/s. Neglecting friction between child and ice, what is the speed (in m/s) of the child after throwing the stone?
I'm kind of confused about the masses..
I'm kind of confused about the masses..
Answers
Answered by
drwls
What confuses you about the masses?
They they you what they are.
Total momentum is conserved and remains zero.
That means the stone and the child, after the stone is thrown, have equal and opposite momenta.
45*Vchild = -0.6*Vstone = -3.48 kg m/s
Vchild = -3.48/45 = -____ m/s
The minus sign means the child goes backwards, relative to the thrown stone direction. Since that only ask for the speed, you can forget about the minus sgn.
They they you what they are.
Total momentum is conserved and remains zero.
That means the stone and the child, after the stone is thrown, have equal and opposite momenta.
45*Vchild = -0.6*Vstone = -3.48 kg m/s
Vchild = -3.48/45 = -____ m/s
The minus sign means the child goes backwards, relative to the thrown stone direction. Since that only ask for the speed, you can forget about the minus sgn.
Answered by
help
i read the question wrong..thank for the help
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