1. %u2013 /2 points CJ9 7.P.001. My Notes | 1. %u2013 /2 points CJ9 7.P.001. My

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1.%u2013/2 pointsCJ9 7.P.001.My Notes  | 1.%u2013/2 pointsCJ9 7.P.001.My Notes  | Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used A 42-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.2 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her).
magnitude N direction ---Select---opposite the velocity of the skaterthe same direction as the velocity of the skaterupwardnot enough information to tell
(a) What is the velocity of the first log just before the lumberjack jumps off? (Indicate the direction of the velocity by the sign of your answers.)
m/s

(b) Determine the velocity of the second log if the lumberjack comes to rest on it.
m/s
Question Part Points Submissions Used

Explanation / Answer

Number 1

Apply Ft = mv

F(.8) = (42)(1.2)

F = 63 N in the opposite direction of the velocity

Number 2a)

By conservation of momentum

mv = mv

(98)(3.3) = (285)v

v = -1.13 m/s (Negative since it will go in the opposite direction)

Number 2b)

Apply conservation of momentum

mv = (m + m)(v)

(98)(3.3) = (98 + 285)(v)

v = .844 m/s

Number 3a)

Ft = mv

F(8 X 10^-3) = (.18)(66)

F = 1485 N (opposite the direction of the puck)

You may have to enter -1485N for this answer since the force is opposite the direction of the puck

Number 3b)

F(8 X 10^-3) = (.18)(66)(2)

F = 2970 N (opposite the original direction of the puck)

You may have to enter -2970 N for this answer since the force is opposite the original direction of the puck

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