{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# a7_001 - After the collision the two skaters move as a unit...

This preview shows page 1. Sign up to view the full content.

PHY 2053, Section 3794, Fall 2009, Quiz 7 1. A 0.450 kg steel ball strikes a wall with a speed of 12.0 m/s at an angle of 60 . 0 with the surface. It remains in contact with the wall for 0.150 s and bounces off with the same speed and angle. a) What is the average force exerted on the ball by the wall (magnitude and direction)? Momentum changes only in the x -direction. ∆ p x = 2 mv cos( θ ) = ¯ F t . We find: ¯ F = 2 mv sin( θ ) t = 2 × 0 . 450 × 12 . 0 × sin(60 ) 0 . 150 = 62 . 4 N . So the average force is of magnitude 62.4 N in the negative x -direction. b) What is the total work done by the ball on the wall? Since the speed doesn’t change, ∆ K.E. = 0, so no net work is done on the wall; i.e., W net = 0 . 2. A 63.0 kg ice skater moving at 12.0 m/s crashes into a stationary skater of unknown mass.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: After the collision the two skaters move as a unit at 4.0 m/s. The impact time is 0.140 s. a) What is the magnitude of the average force experienced by each skater? Using ¯ F ∆ t = m ∆ v , we ﬁnd for the ﬁrst skater: ¯ F = m ∆ v ∆ t = 63 . × − 8 . . 140 = − 3600 N . The forces on the two skaters are equal and opposite. They are both of magnitude 3600 N. b) What is the mass of the second skater? Using ¯ F ∆ t = m ∆ v for the second skater, we ﬁnd: m = ¯ F ∆ t ∆ v = 3600 × . 14 4 . = 126 kg ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online