Question

A girl slides down a snowy hill on an inflated tube sled where she collides with a boy who is stationary in another tube sled. After the collision the girl moves off at an angle theta 1 =0.35 radians from her original direction of motion and the girl moves off at 1.5 times the boy’s final speed . The girl and her sled have half the mass of the boy and his sled. At what angle theta 2 radians does the boy move after the collision ? The angles theta 1 and theta 2 not drawn to scale

Answer 1

m G mla GP 1.5 V B) - B U-V 50.30.35 102=8 B U=0 The all given information is shown in figure By consertation of Linear momentum Horizontal component, me Vig +'mxo = m X (1050) 636-33 g + m X VR X Cos og 0.5 VG O. TO 5 VB + Vo Cos Oq Vertical, oto= g X11.543) Sin 10:35) – mox som (0) 2 Sinoz O 0.25 717 Oq = sin (0.25717) Og 0.26 radian a