Answer:
The final velocity of the first glider is 3.39553 m/s in the opposite direction
The final velocity of the second glider is 0.31553 m/s in the same direction
Explanation:
[tex]m_1[/tex] = Mass of first glider = 0.14 kg
[tex]m_2[/tex] = Mass of second glider = 0.299 kg
[tex]u_1[/tex] = Initial Velocity of first glider = 0.8 m/s
[tex]u_2[/tex] = Initial Velocity of second glider = -2.28 m/s
[tex]v_1[/tex] = Final Velocity of first glider
[tex]v_2[/tex] = Final Velocity of second glider
As momentum and Energy is conserved
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}[/tex]
From the two equations we get
[tex]v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.14-0.299}{0.14+0.299}\times 0.8+\frac{2\times 0.299}{0.14+0.299}\times -2.28\\\Rightarrow v_1=-3.39553\ m/s[/tex]
The final velocity of the first glider is 3.39553 m/s in the opposite direction
[tex]v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.14}{0.14+0.299}\times 0.8+\frac{0.299-0.14}{0.14+0.299}\times -2.28\\\Rightarrow v_2=-0.31553\ m/s[/tex]
The final velocity of the second glider is 0.31553 m/s in the same direction
Answer:
v1 = 2.76 m/s and v2 = - 0.32 m/s
Explanation:
m1 = 0.140 kg
m2 = 0.299 kg
u1 = 0.80 m/s
u2 = - 2.28 m/s
Let the speed after collision is v1 and v2.
Use conservation of momentum
m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2
0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2
0.112 - 0.68 = 0.14 v1 + 0.299 v2
0.14 v1 + 0.299 v2 = - 0.568 ..... (1)
By the use of coefficient of restitution, the value of e = 1 for elastic collision
[tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]
u2 - u1 = v1 - v2
- 2.28 - 0.8 = v1 - v2
v1 - v2 = 3.08
v1 = 3.08 + v2
Put in equation (1)
0.14 (3.08 + v2) + 0.299 v2 = - 0.568
0.43 + 0.44 v2 = - 0.568
v2 = - 0.32 m/s
and
v1 = 3.08 - 0.32 = 2.76 m/s
Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s
