What Is the Speed of a Ball After Bouncing Off a Charging Elephant?

[vortec135]

elastic

a charging bull elephant with mass of 5230 kg comes directly towards you with a speed of 4.45 m/s. you toss a 0.150 kg rubber ball at the elephant with a speed of 7.91 m/s. When the ball bounces back to you what is its speed?

not too sure about how to approach this question.. thanks!

 

[alphysicist]

Hi vortec135,

elastic

a charging bull elephant with mass of 5230 kg comes directly towards you with a speed of 4.45 m/s. you toss a 0.150 kg rubber ball at the elephant with a speed of 7.91 m/s. When the ball bounces back to you what is its speed?

not too sure about how to approach this question.. thanks!

You've said it is an elastic collision. What is conserved in an elastic collision? What two equations are used in this case?

 

[thomate1]

I would approach this question by first understanding the concept of momentum and how it relates to the collision between the bull and the ball. Momentum is defined as the product of an object's mass and its velocity, and it is conserved in a closed system (where no external forces act on the system). In this scenario, the bull and the ball can be considered as a closed system.

The initial momentum of the system can be calculated by multiplying the mass of the bull (5230 kg) by its initial speed (4.45 m/s). This gives us a value of 23258.5 kg*m/s. Since there are no external forces acting on the system, the conservation of momentum tells us that the final momentum of the system must also be 23258.5 kg*m/s.

Now, we can consider the collision between the bull and the ball. This is an elastic collision, which means that both momentum and kinetic energy are conserved. The ball, with a mass of 0.150 kg, has an initial momentum of (0.150 kg)*(7.91 m/s) = 1.1865 kg*m/s. After the collision, the ball bounces back with a speed that we need to calculate.

To do this, we can use the equation for conservation of momentum:

Initial momentum = final momentum

(0.150 kg)*(7.91 m/s) = (0.150 kg)*(v final)

Solving for v final, we get:

v final = (0.150 kg)*(7.91 m/s) / (0.150 kg) = 7.91 m/s

Therefore, the final speed of the ball after it bounces back from the bull is 7.91 m/s. This is the same as its initial speed, as expected for an elastic collision.

In conclusion, the speed of the ball after it bounces back from the bull is 7.91 m/s. This can be explained by the conservation of momentum and the fact that the collision between the bull and the ball was elastic.