A 5.0 kg shell is fired vertically upward. At its max height of 350.0 m, it explodes into 3
pieces. A 1.0 kg chunk lands 15.0 m away, 45
O
due northeast from the launch site. A second 1.0
kg piece lands 25.0 m, 60
O
south of east from the launch site. If all 3 pieces land at the same time
on the (level) ground, where does the third piece land?
Since the shell was at the highest point, its velocity=0. The chunks moved in horizontal plane. Actually, each velocity is horizontal velocity, therefore, these are the velocities of uniform motions.
v1•t=s1=15, v2•t=s2=25 => v1=0.6•v2,
p1=m1•v1 =m1•0.6•v2,
p2=m2•v2
Taking into account the directions of the chunks motions, let’s find the vecor sum of p1 and p2. The angle that p1 and p2 make is 45+60=105º, => (360 - 2•105) =75º. Using the cosine law,
p12= sqrt{p1²+p2²-2p1•p2•cosα} =
=sqrt{ {(m1•0.6•v2)² +(m2•v2)² -2m1•m2•v2•0.6•v2•cos75º}.
||m1=m2||
p12= m2•v2•sqrt{0.6²+1 - 2•0.6•0.26} =1.02•m2•v2
According to the law of conservation of linear momentum
p3 = p12
p3=m3•v3 =>
m3•v3 =1.02•m2•v2,
v3=1.02•m2•v2/m3
since m2=1 kg, m3=5-1-1=3 kg,
v2=25/t
v3= =1.02•1•v2/3=1.02•25/3•t.
s3=v3•t= 1.02•25•t/3•t=1.02•25/3=8.5 m
To find the landing position of the third piece, we can break down the problem into horizontal and vertical components.
First, let's analyze the horizontal motion of the shell. From the information given, we know that the first piece lands 15.0 m away and is at an angle of 45° northeast from the launch site. This means that the first piece has moved 15.0 m horizontally in the positive x-direction.
Next, let's analyze the vertical motion of the shell. The shell reaches its maximum height of 350.0 m above the launch site. At this point, it explodes into three pieces and starts to fall back to the ground. Since the problem does not provide the exact time when the explosion occurs, we can assume that the time it takes for the shell to reach the maximum height is the same as the time it takes for the three pieces to fall back to the ground.
Using the equation for vertical motion, we can find the time taken for the shell to reach the maximum height:
v_final = v_initial + at,
where v_final = 0 m/s (since the shell reaches its maximum height and stops momentarily), v_initial = unknown, a = -9.8 m/s^2 (acceleration due to gravity), and t = unknown.
Simplifying the equation, we have:
0 = v_initial - 9.8t.
We can also use the equation for position in vertical motion:
y = y_initial + v_initialt + 0.5at^2,
where y = 350.0 m (maximum height), y_initial = 0 m (launch site), v_initial = unknown, a = -9.8 m/s^2 (acceleration due to gravity), and t = unknown.
Substituting the known values, we get:
350.0 = 0 + v_initialt - 0.5(9.8)t^2.
Now we have two equations with two unknowns: v_initial and t. By solving these equations simultaneously, we can find the time it takes for the shell to reach the maximum height and the initial velocity of the shell.
Once we have the value of t, we can use it to find the horizontal displacement of the third piece. The third piece is a 1.0 kg chunk, so we can assume the horizontal and vertical components follow the same motion laws as the shell. We already know the velocity in the x-direction is constant and equal to the initial velocity of the shell. Using this initial velocity and the time t, we can calculate the horizontal displacement of the third piece.
I hope this explanation helps you understand the process of solving the problem. It involves breaking down the problem into horizontal and vertical components, applying the laws of motion, and solving simultaneous equations to find the necessary values.