Question

A group of students are performing an experiment in class using collision carts on a track. An 8.0 kg cart moving to the right at 4.0 m/s hits a 4.0 kg cart moving to the left at 6.0 m/s. Immediately after the collision, the 4.0 kg cart moves to the right at 3.0 m/s. What are the speed and direction of the 8.0 kg cart after the collision?

Answer 1

Given that,

Mass of cart = 8.0 kg

Speed of cart = 4.0 m/s in right

Mass of another cart = 4.0 kg

Speed of another cart = 6.0 m/s in left

Speed of another cart after collision = 3.0 m/s in right

We need to calculate the speed and direction of the 8.0 kg cart after the collision

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Put the value into the formula

$8.0\times(4.0)+4.0\times(-6.0)=4.0\times3.0+8.0\times v$

$v=\dfrac{8.0\times4.0-4.0\times6.0-4.0\times3.0}{8.0}$

$v=-0.5\ m/s$

Negative sign shows the direction of motion of the 8.0 kg cart

Hence, The speed and direction of the 8.0 kg cart after the collision is 0.5 m/s in left.

Answer 2

Explanation:

Given that,

Mass of cart = 8.0 kg

Speed of cart = 4.0 m/s in right

Mass of another cart = 4.0 kg

Speed of another cart = 6.0 m/s in left

Speed of another cart after collision = 3.0 m/s in right

We need to calculate the speed and direction of the 8.0 kg cart after the collision

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}m

1

u

1

+m

2

u

2

=m

1

v

1

+m

2

v

2

Put the value into the formula

8.0\times(4.0)+4.0\times(-6.0)=4.0\times3.0+8.0\times v8.0×(4.0)+4.0×(−6.0)=4.0×3.0+8.0×v

v=\dfrac{8.0\times4.0-4.0\times6.0-4.0\times3.0}{8.0}v=

8.0

8.0×4.0−4.0×6.0−4.0×3.0

v=-0.5\ m/sv=−0.5 m/s

Negative sign shows the direction of motion of the 8.0 kg cart

Hence, The speed and direction of the 8.0 kg cart after the collision is 0.5 m/s in left.