Question

Just before they collide,two disk on a horizontal surface have velocities shown In figure.
Knowing that 90 N disk A rebounds to the left with a velocity of 1.8 m/s.Determine the rebound velocity of the 135 N disk B.Assume the impact is perfectly elastic.

Answer 1

Given : W_A = 90N
W_B = 135 N
Taking velocity direction towards right as positive and towards left as negative
Initial velocity of disk A= 3.6 m/s
Final velocity of disk A=-1.8 m/s
Initial velocity of disk B= 3 m/s
To find : Rebound velocity of disk B
Solution :

`m_A=90/gkg`
`m_B=135/gkg`

Consider the X and Y components of uB
u_BX = -u_Bcos35 = -2.4575 m/s
u_BY = -u_Bsin35 = -1.7207 m/s
APPLYING LAW OF CONSERVATION OF MOMENTUM :
m_Au_A + m_Bu_B = m_Av_A + m_Bv_B
`90/gx3.6+135/gx(-2.4575)=90/gx((-1.8)+135/gxv_(BX)`
v_BX = 1.1425 m/s
As the impact takes place along X-axis,the velocities of two disks remains same along Y-axis
v_BY = u_BY = -1.7207 m/s

`V=sqrt((V_(BX))^2+(v_(BY))^2)`
`V=sqrt(1.1425^2+(-1.7207)^2)`
`v = 2.0655 m//s`
`alpha = tan^-1((-1,7207)/(1,1425))`
α = 56.4169°
VELOCITY OF DISK B AFTER IMPACT = 2.0655 m/s (56.4169o in fourth quadrant)