This is an elastic collision problem that involves an orange puck and a green pu

Question

This is an elastic collision problem that involves an orange puck and a green puck on a frictionless air hockey table. Before the collision, the orange puck of mass 0.5 kg is moving with velocity 4.00 m/s along the x-axis. The green puck has mass 0.3 kg and is originally at rest. After the collision, the orange puck as a final velocity of velocity of 2 m/s in an unknown direction "A " degrees above the x-axis. The green puck moves off in an unknown direction "B" degrees below the x-axis. What is angle "B" ? Round your answer to the nearest 0.1 degrees.

Explanation / Answer

m1=0.5kg, m2=0.3kg, v1i=4.0m/s,v2i=0m/s, v1f=2m/s

By law of conservation of linear momentum,

m1v1i+m2v2i=m1v1f+m2v2f

Along x axis,

m1v1ix+m2v2ix=m1v1fx+m2v2fx

0.5*4.0 = 0.5*v1fcosA + 0.3*v2fcosB

0.5*4.0+0.3*0 = 0.5*2*cosA + 0.3*v2fcosB -----------------(1)

Along y axis,

m1v1iy+m2v2iy=m1v1fy+m2v2fy

0.5*4.0 = 0.5*v1fsinA + 0.3*v2fsinB

0.5*0+0.3*0 = 0.5*2*sinA + 0.3*v2fsinB -----------------(2)

From (2),

0=0.5*2*sinA + 0.3*v2fsinB

0=sinA +0.3*v2fsinB

sinA= (-v2fsinB)/0.3 -----------------(3)

Put (3) in (1),

0.5*4.0+0.3*0 = 0.5*2*[-(v2fsinB)/0.3]+ 0.3*v2fsinB

Gives, v2fsinB= -0.6593 m/s --------------(4)

Plug (4) in (2)

0.5*0+0.3*0 = 0.5*2*sinA + 0.3*-0.66

sinA=0.1978

A= sin^-1(0.1978)

A= 11.41 deg --------------(5)

Plug (5) in (1)

0.5*4.0+0.3*0 = 0.5*2*cos11.41 + 0.3*v2fcosB

v2fcosB = 3.4    ---------------(6)

From (4) and (4)

tanB = v2fsinB ()/(v2fcosB) = -0.6593/3.4

B= tan^-1(-0.6593/3.4)

B = -10.97 = 11.00 deg

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