HW_10_Solution_P1112_F08

HW_10_Solution_P1112_F08 - Name: _ Physics 1112, Fall 2008,...

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

View Full Document Right Arrow Icon
Name: __________________________________ Physics 1112, Fall 2008, Assignment 10 Solutions Cornell University 1 Physics 1112, Fall 2008, Assignment 10 Solutions 1. Let the x + direction be east and let the y + direction be north. Mass of car: kg 1400 = m . Mass of truck: kg 2800 = M . Initial velocity of car: i v ˆ ) km/h 0 . 35 ( i ! = m r . Initial velocity of truck: j v ˆ ) km/h 0 . 50 ( i ! = M r . x m x mv p i i = . y M y Mv p i i = . x x v M m p f f ) ( + = . y y v M m p f f ) ( + = . Because no net external force acts on the system consisting of the two vehicles, momentum is conserved: x x m v M m mv f i ) ( + = . y y M v M m Mv f i ) ( + = . ) ( i f M m mv v x m x + = . ) ( i f M m Mv v y M y + = . kg 4200 ) km/h 0 . 50 ( ) kg 2800 ( ) km/h 0 . 35 ( ) kg 1400 ( ) ( 2 2 2 2 2 i 2 2 i 2 2 f 2 f f ! + ! = + + = + = M m v M v m v v v y M x m y x km/h 3 . 35 f = v . ° = ! ! " # $ $ % = ! ! " # $ $ % = ! ! " # $ $ % = 7 . 70 ) 0 . 35 )( 1400 ( ) 0 . 50 )( 2800 ( tan tan tan 1 i i 1 f f 1 x m y M x y mv Mv v v ( . So the vehicles are moving at km/h 3 . 35 f = v in the direction 70.7 ° south of west. 2. Let the x + direction be the direction of travel of the bullet just before it hits the block. Mass of bullet: kg 10 00 . 5 g 00 . 5 3 ! " = = m . Mass of block: kg 20 . 1 = M . 20 . 0 k = μ . Distance the block with bullet slides: m 230 . 0 = d . The strategy here is to use conservation of momentum to get the speed of the block & bullet just after the bullet has embedded in the block and then to relate this to the distance that the block with bullet slides via work-energy considerations. The block will start to move as soon as the bullet touches it and so kinetic friction will then be a net external force on the system consisting of the block and bullet during the collision as the bullet is embedding in the block. Strictly speaking then, momentum of the bullet-block system is not conserved during this collision. However, the duration of the collision is very short and the kinetic friction force magnitude is small compared with the magnitudes of the average forces that the bullet and block exert on each other during the collision. Therefore we can apply conservation of momentum as a good approximation: bullet i mv p x = . f
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/28/2009 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

Page1 / 5

HW_10_Solution_P1112_F08 - Name: _ Physics 1112, Fall 2008,...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online