Question

A 0.140-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.299-kg glider that is moving to the left with a speed of 2.28 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

Answer 1

v1 = 2.76 m/s and v2 = - 0.32 m/s

Step-by-step 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

`e=(v_(1)-v_(2))/(u_(2)-u_(1))`

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

Answer 2

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

Step-by-step explanation:

`m_1` = Mass of first glider = 0.14 kg

`m_2` = Mass of second glider = 0.299 kg

`u_1` = Initial Velocity of first glider = 0.8 m/s

`u_2` = Initial Velocity of second glider = -2.28 m/s

`v_1` = Final Velocity of first glider

`v_2` = Final Velocity of second glider

As momentum and Energy is conserved

`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`

`{\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)`

From the two equations we get

`v_(1)=(m_1-m_2)/(m_1+m_2)u_(1)+(2m_2)/(m_1+m_2)u_2\\\Rightarrow v_1=(0.14-0.299)/(0.14+0.299)* 0.8+(2* 0.299)/(0.14+0.299)* -2.28\\\Rightarrow v_1=-3.39553\ m/s`

The final velocity of the first glider is 3.39553 m/s in the opposite direction

`v_(2)=(2m_1)/(m_1+m_2)u_(1)+(m_2-m_1)/(m_1+m_2)u_2\\\Rightarrow v_2=(2* 0.14)/(0.14+0.299)* 0.8+(0.299-0.14)/(0.14+0.299)* -2.28\\\Rightarrow v_2=-0.31553\ m/s`

The final velocity of the second glider is 0.31553 m/s in the same direction