Question

Can someone work this problem out completly?

A 3 kg model rocket is launched vertically and reaches
andaltitude of 60 m wih a speed of 28 m/s at the end of flight,
timet=0. As the rocket approaches its maximum altitude itexplodes
into two parts of masses Ma = 1 kg and Mb = 2 kg. Part A is
observed to strike the ground 74.4 m west of the launchpoint at t =
5.85 s. Determine the position of part B at thattime.

Answer 1

Part B will be 37.2 m to the east of the launch point at an altitude of 80.8 m

Step-by-step explanation:

At t0 the rocket is at 60 m and has a speed of 28 m/s.

Then it explodes when it reaches max altitude.

Since the rocket will have its engine off after t0 it will be in free fall.

It will be affected only by the acceleration of gravity, so it moves with constant acceleration, we can use this equation.

Y(t) = Y0 + V0*t + 1/2*a*t^2

Y0 = 60 m

V0 - 28 m/s

a = g = -9.81

We also have the equation for speed:

V(t) = V0 + a*t

And we know that when it reaches its highest point it ill have a speed of zero.

0 = V0 + a * t

a * t = -V0

t = -V0 / a

t = -28 / -9.81 = 2.85 s

This is the time after t0 when the engine ran of of fuel.

Using this value on the position equation:

Y(2.85) = 60 + 28*2.85 + 1/2*(-9.81)*2.85^2 = 100 m

At an altitude of 100 m it explodes in two parts. An explosion is like a plastic collision in reverse, momentum is conserved.

Since the speed of the rocket is zero at that point, the total momentum will be zero too.

Part A, with a mass of 1 kg fell 74.4 m west of the launch point. It fell in a free fall with a certain initial speed given to it by the explosion. This initial speed had vertical and horizontal components.

First the horizontal. In the horizontal there is no acceleration (ignoring aerodynamic drag).

X(t) = X1 + Vx1 * (t - t1)

X1 = 0 because it was right aboce the launch point. t1 is the moment of the explosion.

t1 = 2.85 s

We know it fell to the ground at t = 5.85 s

Vx1 * (t2 - t1) = X(t2) - X1

Vx1 = (X(t2) - X1) / (t2 - t1)

Vx1 = (74.4 - 0) / (5.85 - 2.85) = 24.8 m/s

Now the vertical speed

Y(t2) = Y1 + Vy1*(t2 - t1) + 1/2*a*(t2 - t1)^2

Vy1*(t2 - t1) = Y(t2) - 1/2*a*(t2 - t1)^2 - Y1

Y(t2) = 0 because it is on the ground.

Vy1 = (-1/2*a*(t2 - t1)^2 - Y1) / (t2 - t1)

Vy1 = (-1/2*-9.81*(5.85 - 2.85)^2 - 100) / (5.85 - 2.85) = -18.6 m/s

If the part A had a speed of (24.8*i - 18.6*j) m/s, we calcultate the speed of part 2

Horizontal:

vxA * mA + vxB * mB = 0

-vxA * mA = vxB * mB

vxB = -vxA * mA / mB

vxB = -24.8 * 1 / 2 = -12.4 m/s

Vertical:

vyA * mA + vyB * mB = 0

-vyA * mA = vyB * mB

vyB = -vyA * mA / mB

vyB = -(-16.6) * 1 / 2 = 8.3 m/s

Now that we know its speed and position at t1 we can know ehre it will be at t2

X(t) = 0 - 12.4 * (t - t1)

X(5.85) = -12.4 * (5.85 - 2.85) = -37.2 m

Y(t) = 100 + 8.3 * (t - t1) + 1/2 * (-9.8) * (t - t1)^2

Y(t) = 100 + 8.3 * (5.85 - 2.85) + 4.9 * (5.85 - 2.85)^2 = 80.8 m

Part B will be 37.2 m to the east of the launch point at an altitude of 80.8 m