# H_Chapter_7_answer_Problems_students - Chapter 7 Homework...

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Chapter 7 - Homework - Solution to Problems 5, 7, 14,18,20, 27, 40, 44, 49, 52 ******************************** 5/A system consists of three particles with these masses and velocities: mass 3.0 kg, moving north at 3.0 m/s; mass 4.0 kg, moving south at 5.0 m/s; and mass 7.0 kg, moving north at 2.0 m/s. What is the total momentum of the system? Given: m 1 =3.0kg, v 1 =3.0m/s north; m2=4.0kg, v 2 =5.0m/s south; m3=7.0kg, v 3 =2.0m/s north Question p tot ? north s m kg s m kg s m kg s m kg p north v m v m v m north v m south v m north v m v m v m v m p p p p tot toh / 3 / 0 . 2 0 . 7 / 0 . 5 0 . 4 / 3 0 . 3 ) ( 3 3 2 2 1 1 3 3 2 2 1 1 3 3 2 2 1 1 3 2 1 7/ A ball of mass 5.0 kg moving with a speed of 2.0 m/s in the +x-direction hits a wall and bounces back with the same speed in the –x-direction. What is the change of momentum of the ball? Given: m=5.0kg v xi =+2.0m/s , v xf =-2 .0m/s Question: p ? Remark : The vector quantities are written either with an arrow or with a bold character. The change in momentum is p = p f p i = m v f m v i = m( v f v i ) = (5.0kg) (- 2.0m/s 2.0m/s) = 20kg m/s in the x direction 14 / A 3 .0 - kg body is initially moving northward at 15 m/s. Then a force of 15 N, toward the east, acts on it for 4.0 s. (a) At the end of the 4.0 s, what is the body’s final velocity? (b) What is the change in momentum during the 4.0 s? The change of momentum i n each direction is: p N orth =0; p E ast = F av t = m v E ast = mv E ast The magnitude of v in the final direction is V f = sqrt(v North 2 + v East 2 ) = sqrt(v North 2 + (F av t/m) 2 ) = sqrt( (15m/s2) + (15N) ( (4.0s)/(3.0kg) ) 2 ) = 25 m/s T = tan -1 (v North /v

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