Question

0/1 points 1 Previous Answers KatzPSEf1 11. P025. My l A 2.80-kg ball is shot into a 0.470-kg box that is at rest on a frictionless, horizontal table (see figure below): after the collision, the bal is embedded in the e, the bar attached to a 0.30-m rope that is attached to the table on the other end. The ball's initial velocity is perpendicular to the rope as shown. If the ball's initial speed before impac is 13.0 m/s, what is the tension in the rope after the collision? 153.363 X Your response differs from the correct answer by more than 10%. Double check your calculations. 0.30m Need Help? Resd

Answer 1

Initial angular momentum = 2.8 times 13 times 0.3 = 10.92 kg m/s final momentum = (2.8 + 0.47) v times 0.3 So, (2.8 + 0.47)v times 0.3 = 10.92 implies v = 36.4/3.27 = 11.13 m/sec Tension = mv^2/r = (2.8 + 0.47) (11.13)^2/0.3 = 1350N initial momentum is x-direction = m_p times 1.92 times 10^8 Final momentum in x-direction = m_p times (1.646 times 10^8)cos 31 degree + m_p times v cos theta. therefore 1.92 times 10^8 = 1.41 times 10^8 + v cos theta ...I Also, for y-axis. 1.646 times 10^8 sin 31 degree - v sin theta = 0. implies v sin theta = 0.84 times 10^8 II From 1 and II v^2 = (1.92 - 1.41)^2 times 10^16 + (0.84)^2 times 10^16 v = 0.98 times 10^8 and theta = sin - 0.84/0.98 = 59 degree.