Solutions to Chapter 6 Review Problems

# Therefore the third fragment must fly off toward a

• Notes
• DrRockNarwhal5987
• 4

This preview shows pages 2–4. Sign up to view the full content.

Therefore, the third fragment must fly off toward a general direction of (2) north of east . (b) Apply momentum conservation P  o = P  . along the x axis: (2.0 kg)(0) = (0.50 kg)( 2.8 m/s) + (1.3 kg)(0) + (1.2 kg) v x , so v x = 1.17 m/s; along the y axis: (3.0 kg)(0) = (0.50 kg)(0) + (1.3 kg)( 1.5 m/s) + (1.2 kg) v y , so v y = 1.63 m/s. v = (1.17 m/s) 2 + (1.63 m/s) 2 = 2.0 m/s , θ = tan 1 163 117 . . = 54 ° north of east . 41. (a) Energy conservation gives mgh = 1 2 mv 2 v = [2(9.80 m/s 2 )(0.10 m)] 1/2 = 1.4 m/s (b) The balls stick together, so their speed just after the collision is 0.700 m/s, by momentum conservation. Using energy conservation again gives h = v 2 /2 g = (0.700 m/s) 2 /[2(9.80 m/s 2 )] = 0.025 m = 2.5 cm 46. m 1 = 0.010 kg, m 2 = 3.0 kg, v 1o = 400 m/s, v 2o = 0, v 1 = 300 m/s, v 2 = ? P  o = P  , m 1 v 1o + m 2 v 2o = m 1 v 1 + m 2 v 2 . v 2 = m 1 v 1o + m 2 v 2o m 1 v 1 m 2 = (0.010 kg)(400 m/s) + 0 (0.010 kg)(300 m/s) 3.0 kg = 0.33 m/s . y 3 1 x 2

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