Question

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected 30 degrees from the horizontal and the 3kg object is deflected 315 degrees from the horizontal. Determine the velocity of each ball after the collision.

Answer 1

The velocity of the combined object after the collision is 1.875 m/s. The final velocity of both the 5kg object and the 3kg object is 1.875 m/s.

Step-by-step explanation:

In this problem, we can use the principles of conservation of momentum and conservation of kinetic energy to determine the velocities of the two objects after the collision.

First, we need to calculate the initial momentum of each object. The momentum is the product of the mass and velocity of an object. For the 5kg object, its initial momentum is 5kg * 3m/s = 15 kg*m/s, and for the 3kg object, its initial momentum is 3kg * 0m/s = 0 kg*m/s.

After the collision, the two objects stick together and move as one. Let's denote the final velocity of the combined object as V. According to the conservation of momentum, the initial momentum of the system is equal to the final momentum of the system. Therefore, we have:

Initial momentum of the system = final momentum of the system

15 kg*m/s + 0 kg*m/s = (5kg + 3kg) * V

15 kg*m/s = 8kg * V

V = 15 kg*m/s / 8kg = 1.875 m/s

So the velocity of the combined object after the collision is 1.875 m/s.

Since the two objects move as one, the final velocity of the 5kg object is the same as the final velocity of the combined object, which is 1.875 m/s.

Now, let's determine the final velocity of the 3kg object after the collision. Since the two objects are sticking together and moving as one, their final velocity is the same. Therefore, the final velocity of the 3kg object is also 1.875 m/s.

Answer 2

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Step-by-step explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)\cos\theta+m_(2)v_(2)\cos\theta`

Put the value into the fomrula

`5*3+0=5* v_(1)\cos30+3* v_(2)\cos45`

`15=5v_(1)*(√(3))/(2)+3v_(2)*(1)/(√(2))`

`15=(5√(3))/(2)v_(1)+(3)/(√(2))v_(2)`....(I)

Along y -axis

`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)\sin\theta+m_(2)v_(2)\sin\theta`

Put the value into the formula

`0+0=5* v_(1)\sin30-3* v_(2)\sin45`

`(5)/(2)v_(1)-(3)/(√(2))v_(2)=0`...(II)

From equation (I) and (II)

`v_(1)=(15*2)/(5√(3)+5)`

`v_(1)=2.19\ m/s`

Put the value of v₁ in equation (I)

`(5)/(2)*2.19-(3)/(√(2))v_(2)=0`

`v_(2)=(5.475*√(2))/(3)`

`v_(2)=2.58\ m/s`

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.