Question

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. The 1.0-kg part falls straight down after breakup with an initial speed of 10.0 m/s, the 0.7-kg part moves in the original forward direction, and the 0.3-kg part goes straight up.

(a) Find the speeds of the 0.3-kg and 0.7-kg pieces immediately after the break-up.

(b) How high from the break-up point does the 0.3-kg piece go before coming to rest

Answer 1

a)
`v_(0.7kg)=109.43(m)/(s)`
`v_(0.3kg)=33.33(m)/(s)`

b)h=56.68 m

Step-by-step explanation:

1) Notation and basic concepts

Projectile motion

When the motion is in two dimensions is known as projectile motion. The path that follows the object is known as trajectory.

Other important concepts that we will use to solve this problem are Conservation of momentum, this concept is a fundamental law of physics which states "that the momentum of a system is constant if there are no external forces acting on the system"

And the other important concept is conservation of energy a principle of physics that states "the energy of interacting bodies or particles in a closed system remains constant".

`\theta =40` is the initial angle for the motion

`v_i=50(m)/(s)` is the initial velocity

`v_(1kg)=10(m)/(s)` is the velocity for the 1kg object

`v_(0.7kg)` is the velocity for the 0.7kg object (variable of interest)

`v_(0.3kg)` is the velocity for the 0.3kg object (variable of interest)

`h_(0.3kg)` maximum height reached by the 0.3kg object

The figure attached describe the situation for the problem.

2) Part a

The momentum is defined as
`p=mv` using this concept we can find the velocities required for part a

If we use this principle from the starting point and at the highest point for the fragments and we focus on the 0.7 kg, since this fragment follows the same x trajectory we have this

`2kg*50(m)/(s)cos(40)=0.7kgv_(0.7kg)`

And we can solve for
`v_(0.7kg)`, on this case

`v_(0.7kg)=(2kg*50(m)/(s)cos(40))/(0.7kg)=109.43(m)/(s)`

Now for the 0.3 kg part we can use again conservation of momentum this part moves forward, the momentum for this particle upward need to be the same for the particle of 1kg moving downward, so we have the following formula

`1kg*v_(1kg)=0.3kg*v_(0.3kg)`

And solving for
`v_(0.3kg)` we got:

`v_(0.3kg)=(1kg*10(m)/(s))/(0.3kg)=33.33(m)/(s)`

3) Part b

In order to find "How high from the break-up point does the 0.3-kg piece go before coming to rest" we can use conservation of energy when the fragment of 0.3kg begins to move upward we can set up this level as our reference height, at this starting point we have kinetic energy given by
`KE=(1)/(2)mv_(0.3kg)^2`, where
`v_(3kg)` is the velocity founded on part a), and 0 potential energy since we are at the reference level. And we want to find the maximum height h that this part achieves, so at this final point the velocity for the particle would be 0 so we don't have kinetic energy but the potential energy would be maximum and given by
`PE=mgh`, applying the balance of energy we got:

`(1)/(2)mv^2_(0.3kg)=mgh`

Cancelling the mass and solving for h we got

`h=((1)/(2)(33.33(m)/(s))^2)/(9.8(m)/(s^2))=56.68m`