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# UPTOSAL (5 points extra credit) How fast is the box moving after it ha.docx

UPTOSAL (5 points extra credit) How fast is the box moving after it has slid 1.00 m? 3. (10 points) Two skaters, one weighing 525 N, the other weighing 625 N, are standing next to each other on a frictionless ice rink. The skaters push off from each other and start sliding in opposite directions. If the heavier skater is sliding at 1.5 m/s after the push, how fast will the lighter one be going? 4. (10 points) Pebbles of mass m are thrown at speed v, off the edge ofa cill ofhribt
Solution
This is a conservation of momentum problem.
Initial momentum = 0
Final momentum = 625/9.8 * 1.5 + 525/9.8 * v
Final momentum = initial momentum
625/9.8 * 1.5 + 525/9.8 * v = 0
v = -1.786 m/s
The negative sign means the lighter skate is moving in the opposite direction of the heavier skater.
So the lighter one is going 1.786 m/s fast.
.

UPTOSAL (5 points extra credit) How fast is the box moving after it has slid 1.00 m? 3. (10 points) Two skaters, one weighing 525 N, the other weighing 625 N, are standing next to each other on a frictionless ice rink. The skaters push off from each other and start sliding in opposite directions. If the heavier skater is sliding at 1.5 m/s after the push, how fast will the lighter one be going? 4. (10 points) Pebbles of mass m are thrown at speed v, off the edge ofa cill ofhribt
Solution
This is a conservation of momentum problem.
Initial momentum = 0
Final momentum = 625/9.8 * 1.5 + 525/9.8 * v
Final momentum = initial momentum
625/9.8 * 1.5 + 525/9.8 * v = 0
v = -1.786 m/s
The negative sign means the lighter skate is moving in the opposite direction of the heavier skater.
So the lighter one is going 1.786 m/s fast.
.

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### UPTOSAL (5 points extra credit) How fast is the box moving after it ha.docx

1. 1. UPTOSAL (5 points extra credit) How fast is the box moving after it has slid 1.00 m? 3. (10 points) Two skaters, one weighing 525 N, the other weighing 625 N, are standing next to each other on a frictionless ice rink. The skaters push off from each other and start sliding in opposite directions. If the heavier skater is sliding at 1.5 m/s after the push, how fast will the lighter one be going? 4. (10 points) Pebbles of mass m are thrown at speed v, off the edge ofa cill ofhribt Solution This is a conservation of momentum problem. Initial momentum = 0 Final momentum = 625/9.8 * 1.5 + 525/9.8 * v Final momentum = initial momentum 625/9.8 * 1.5 + 525/9.8 * v = 0 v = -1.786 m/s The negative sign means the lighter skate is moving in the opposite direction of the heavier skater. So the lighter one is going 1.786 m/s fast.