I have the answers, I just need help getting to the answer. Please and thank you
ID: 1387602 • Letter: I
Question
I have the answers, I just need help getting to the answer. Please and thank you!
Two blocks of clay slide over a level, frictionless plane. Block A has a mass of 5.00 kg and slides due east at an initial speed of 10.0 m/s. Block B has a mass of 1.5 kg and slides in a direction 30.0 degrees west of due north at an initial speed of 20.0 m/s. They collide and stick together. (a) Calculate the final speed of the resulting blob. (b) Calculate the direction of motion of the final blob as an angle north of due east. (Answers: 6.70 m/s, 36.6 degrees)
Explanation / Answer
Let east be +x axis
given, m1 = 5 kg
u1 = 10 m/s
u1 = 10 m/s, u1y = 0
m2 = 1.5 kg,
u2 = 20 m/s
u2x = -20*sin(30) = -10 m/s
u2y = 20*cos(30) = 17.32 m/s
let v is the final speeed of two blocks.
let vx and vy are the componnents of final velocity.
Apply conservation of momentum in x-direction
m1*u1x + m2*u2x = (m1+m2)*vx
==> vx = (m1*u1x + m2*u2x)/(m1+m2)
= (5*10 - 1.5*10)/(5+1.5)
= 5.38 m/s
Apply conservation of momentum in y-direction
m1*u1y + m2*u2y = (m1+m2)*vy
==> vy = (m1*u1y + m2*u2y)/(m1+m2)
= (0 + 1.5*17.32)/(5+1.5)
= 4 m/s
so, v = sqrt(vx^2 + vy^2)
= sqrt(5.38^3 + 4^2)
= 6.7 m/s <<<<-----Answer
Ddrection : theta = tan^-1(vy/vx)
= tan^-1(4/5.38)
= 36.63 degrees north of East <<<<-----Answer