By Norbury J.W.

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**Example text**

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.

Ignore friction and rolling motion and assume the collision is elastic. Also both balls have the same mass. SOLUTION The collision occurs as shown in the figure. We have m1 = m2 ≡ m. y vT vPi m1 m2 α θ x vP 63 Momentum conservation is: pP i = pP + pT and we break this down into the x and y directions. Momentum conservation in the y direction is: 0 = m vT sin α − m vP sin θ vP sin θ = vT sin α Momentum conservation in the x direction is: m vP i = m vT cos α + m vP cos θ vP i = vT cos α + vP cos θ Energy conservation is: 1 1 1 m vP2 i = m vP2 + m vT2 2 2 2 vP2 i = vP2 + vT2 We now have 3 simultaneous equations which can be solved.

A particle of mass m is located on the x axis at the position x = 1 and a particle of mass 2m is located on the y axis at position y = 1 and a third particle of mass m is located off-axis at the position (x, y) = (1, 1). What is the location of the center of mass? SOLUTION The position of the center of mass is rcm = with M ≡ xcm = 1 M 1 M mi ri i mi . The x and y coordinates are i mi xi and ycm = i 1 × m + 2m + m ×(m × 1 + 2m × 0 + m × 1) 1 M mi yi i 1 × m + 2m + m ×(m × 0 + 2m × 1 + m × 1) = = 1 2m (m + 0 + m) = 4m 4m 1 = 2 1 3m (0 + 2m + m) = 4m 4m 3 = 4 = = Thus the coordinates of the center of mass are (xcm , ycm ) = 1 3 , 2 4 59 2.

### Solutions manual for elementary mechanics and thermodynamics by Norbury J.W.

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