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

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       $v_{oy}$ = v₀ sin 40

       $v_{oy}$ = 50.0 sin40

       $v_{oy}$ = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    $p_{oy}$ = 0

After the break

X axis

   $p_{fx}$ = m₂ v₂

Axis y

     $p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       $v_{fy}$² = $v_{oy}$² - 2 gy

       0 =  $v_{oy}$²-2gy

       y =  $v_{oy}$² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      $v_{f}$² =  $v_{y}$² - 2 g y₂

      0 =  $v_{y}$² - 2 g y₂

     y₂ =  $v_{y}$² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

Answer 2

The speeds of the 0.3-kg and 0.7-kg pieces are 33 and 110 m/s respectively.

Given the following data:

• Mass of projectile = 2.0 kg.

• Angle = 40.0°.

• Three masses = 1.0 kg, 0.7 kg, and 0.3 kg.

• Initial speed = 10.0 m/s.

How to calculate their respective speed.

First of all, we would determine the horizontal and vertical component of the initial velocity of the projectiles.

The horizontal component of velocity.

Mathematically, the horizontal component of velocity is given by this formula:

$V_x = Vcos \theta\\\\V_x =50 \times cos40\\\\V_x =50 \times 0.7660$

Vx = 38.30 m/s.

The vertical component of velocity.

Mathematically, the vertical component of velocity is given by this formula:

$V_y = Vsin \theta\\\\V_y =50 \times sin40\\\\V_y =50 \times 0.6428$

Vy = 32.14 m/s.

Next, we would determine the speed of the smaller masses by applying the law of conservation of momentum:

For mass of 0.3 kg:

$m_1v_1=m_3v_3\\\\v_3=\frac{m_1v_1}{m_3} \\\\v_3=\frac{10 \times 1 }{0.3} \\\\v_3=33\;m/s$

For mass of 0.7 kg:

$m_2v_2=mv_y\\\\v_2=\frac{mv_x}{m_2} \\\\v_3=\frac{38.30 \times 2 }{0.7} \\\\v_2=110\;m/s$

b. To determine the height the 0.3 kg mass went before coming to rest, we would apply the third equation of kinematics:

$V^2=U^2+2gy\\\\V^2=0+2gy\\\\y=\frac{V^2}{2g} \\\\y=\frac{33^2}{2\times 9.8} \\\\y=\frac{1089}{19.6}$

y = 56 meters.

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