Question

An object of mass 3M, moving in the +x direction at speed v0, breaks into two pieces of mass M and 2M as shown in the figure below. ?1=47.0?? and ?2=32.0??, determine the final velocities of the resulting pieces in terms of v0.

Determine the x component of the velosity of the smaller piece v1xin terms of v0. (Express your answer to three significant figures.)

Determine the ycomponent of the velosity of the smaller piece v1y in terms of v0. (Express your answer to three significant figures.)

Determine the xcomponent of the velosity of the larger piece v2x in terms of v0. (Express your answer to three significant figures.)

Determine the y component of the velosity of the larger piece v2y in terms of v0. (Express your answer to three significant figures.)

Answer 1

The angle theta_1 = 47 degree and theta_2 = 32 degree the horizonally conservation of momentum 3 MV_0 = 2MV_2 cos theta_2 + MV_1 cos theta_1 3V_0 = 2V_2 cos 32 + V_1 cos 47 degree 3 V_0 = 2V_2 (0.848) + V_1 (0.681) 3 V_0 = V_2 (1.696) + V_1 (0.681) 1 the vertically momentum balance is 2MV_2 sin theta_2 = MV_1 sin theta_1 2V_2 sin 32 degree = V_1 sin 47 degree 2V_2 (0.529) = V_1 (0.731) V_2 (1.058) = V_1 (0.731) V_2 = V_1 (0.69) 2 Substitute 2 in eq 1, we get V_1 3V_0 = V_1 (0.69) (1.696) + V_1 (0.681) V_1 = 3V_0/1.851 = (1.62) V_0 rightarrow V_1 = (1.62) V_0 From 2 rightarrow V_2 (1.62) V_0 times 0.69 = (1.11)V_0 rightarrow V_2 = (1.11)V_0 Now V_1x = V_1 cos 47 degree (1.62)V_0 times 0.681 = (1.1) V_0 V_1y = V_1 sin theta_1 (1.62) V_0 times 0.731 = (1.18) V_0