{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions10

# Solutions10 - Physics 140 Discussion 10 Solutions 8.13...

This preview shows pages 1–2. Sign up to view the full content.

Physics 140 - Discussion 10 - Solutions 8.13. I DENTIFY : The force is constant during the 1.0 ms interval that it acts, so . t = J F r r 2 1 2 1 ( ). m = - = - J p p v v r r r r r S ET U P : Let x + be to the right, so 1 5 00 m/s. x v = + . Only the x component of J r is nonzero, and 2 1 ( ). x x x J m v v = - E XECUTE : (a) The magnitude of the impulse is 3 3 (2 50 10 N)(1 00 10 s) 2 50 N s. J F t = ∆ = . × . × = . The direction of the impulse is the direction of the force. (b) (i) 2 1 . x x x J v v m = + 2 50 N s. x J = + . 2 2 50 N s 5 00 m/s 6 25 m/s. 2 00 kg x v + . = + . = . . The stone’s velocity has magnitude 6.25 m/s and is directed to the right. (ii) Now 2 50 N s x J = - . and 2 2 50 N s 5 00 m/s 3 75 m/s. 2 00 kg x v - . = + . = . . The stone’s velocity has magnitude 3.75 m/s and is directed to the right. E VALUATE : When the force and initial velocity are in the same direction the speed increases and when they are in opposite directions the speed decreases. __________________________________________________________________________________ 8.29. I DENTIFY : The horizontal component of the momentum of the system of the rain and freight car is conserved. S ET U P : Let x + be the direction the car is moving initially. Before it lands in the car the rain has no momentum along the x -axis. E

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Solutions10 - Physics 140 Discussion 10 Solutions 8.13...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online