By Norbury J.W.

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A cannon ball is fired horizontally at a speed v0 from the edge of the top of a cliff of height H. e. the range) that the cannon ball travels. Check that your answer has the correct units. SOLUTION v0 H R In the x (horizontal) direction 1 x − x0 = v0x t + ax t2 2 Now R = x − x0 and ax = 0 and v0x = v0 giving R = v0 t. We obtain t from the y direction 1 y − y0 = v0y t + ay t2 2 Now y0 = 0, y = −H, v0y = 0, ay = −g giving 1 −H = − gt2 2 or t= 2H g Substuting we get R = v0 t = v0 2H g Check units: The units of v0 2H g are √ m −1 = m sec sec2 = m sec−1 sec = m m sec−2 which are the correct units for distance.

FORCE & MOTION - II Substitute for T and N into the left equation F cos θ − m2 a − m2 g − µ(m1 g − F sin θ) = m1 a F (cos θ + µ sin θ) − g(m2 + µm1 ) = m1 a + m2 a a= F (cos θ + µ sin θ) − g(m2 + µm1 ) m1 + m2 45 5. If you whirl an object of mass m at the end of a string in a vertical circle of radius R at constant speed v, derive a formula for the tension in the string at the top and bottom of the circle. SOLUTION T W R T W 46 CHAPTER 5. FORCE & MOTION - II Bottom: ΣFy = may mv 2 T −W = R T T mv 2 R mv 2 = mg + R = W+ Top: ΣFy = may mv 2 T +W = R mv 2 T = −W R mv 2 T = − mg R 47 6.

FORCE & MOTION - II 1. A mass m1 hangs vertically from a string connected to a ceiling. A second mass m2 hangs below m1 with m1 and m2 also connected by another string. Calculate the tension in each string. SOLUTION A) B) T’ T m m 1 2 T’ W m 2 2 Obviously T = W1 +W2 = (m1 +m2 )g. The forces on m2 are indicated in Figure B. Thus Fy = m2 a2y T − W2 = 0 T = W 2 = m2 g 39 2. What is the acceleration of a snow skier sliding down a frictionless ski slope of angle θ ? Check that your answer makes sense for θ = 0o and for θ = 90o .

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