YES LAST PHYSICS QUESTION BEFORE CHRISTMAs!!!!?

Question

please help me get it right!!!


Question 1: An m2 = 1.4 kg can of soup is thrown upward with a velocity of v2 = 5.2 m/s. It is immediately struck from the side by an m1= 0.5 kg rock traveling at v1= 8.2 m/s. The rock ricochets off at an angle of α = 65 degrees with a velocity of v3= 5.4 m/s. What is the angle of the can’s motion after the collision? Answer in units of degrees.

Question 2: With what speed does the can move immediately after the collision? Answer in units of m/s.

Answer 1

I will assume the rock is travelling perfectly horizontally before and after collision
therefore
0.5*8.2=0.5*5.4*cos(65)+1.4*v*cos(θ)
and
0.5*5.4*sin(65)-1.4*v*sin(θ)=0

solve for θ

1.4*v*sin(θ)=0.5*5.4*sin(65)
1.4*v*cos(θ)=0.5*8.2-0.5*5.4*cos(65)

divide

Tan(θ)=5.4*sin(65)/(8.2-5.4*cos(65))
θ=39.6 degrees

that is the angle in the horizontal plane

solving for v in the horizontal plane
vh=2.74 m/s
so the angle from horizontal of the can's motion is
atan(5.2/2.74)

62.2 degrees above horizontal

speed just after collision
sqrt(2.74^2+5.2^2)
5.88 m/s


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