Question

[Computer] A grenade is thrown with initial velocity \(\mathbf{v}_{\mathrm{o}}\) from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: \(\mathbf{v}_{\mathrm{o}}=(4,4), g=1,\) and \(0 \leq t \leq 4\) (and with \(x\) measured horizontally and \(y\) vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at \(t=1,2,3,4 .\) (b) At \(t=4,\) when the grenade's velocity is \(\mathbf{v},\) it explodes into two equal pieces, one of which moves off with velocity \(\mathbf{v}+\Delta \mathbf{v} .\) What is the velocity of the other piece? (c) Assuming that \(\Delta \mathbf{v}=(1,3),\) add to your original plot the paths of the two pieces for \(4 \leq t \leq 9 .\) Insert marks to show their positions at \(t=5,6,7,8,9\). Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.

Short answer

The velocity of the other piece is (3, -3), and the CM follows the original path.

Step-by-step solution

Understand the Initial Conditions

The grenade is initially thrown from the origin with velocity \( \mathbf{v}_o = (4, 4) \) where \( x \) is horizontal and \( y \) is vertical. The acceleration due to gravity is \( g = 1 \). We need to plot its trajectory from \( t=0 \) to \( t=4 \).

Plot the Trajectory of the Grenade

Since gravity affects only the vertical component of velocity, set up the following trajectory equations:- Horizontal position: \( x(t) = v_{ox} \cdot t = 4t \).- Vertical position: \( y(t) = v_{oy} \cdot t - 0.5 \cdot g \cdot t^2 = 4t - 0.5t^2 \).Plot these equations in a suitable program and mark the points at \( t=1, 2, 3, \) and \( 4 \).

Determine the Velocity at t=4

At \( t=4 \), calculate:- \( x(4) = 4 \cdot 4 = 16 \)- \( y(4) = 4 \cdot 4 - 0.5 \cdot 1 \cdot 16 = 16 - 8 = 8 \)The velocity at any time \( t \) is the derivative of the position:- Horizontal velocity \( v_x = 4 \)- Vertical velocity \( v_y = 4 - t \cdot 1 = 0 \) at \( t=4 \)Thus, the velocity is \( \mathbf{v} = (4, 0) \).

Calculate Velocity of the Pieces Post-Explosion

The grenade splits into two pieces with one piece moving at \( \mathbf{v} + \Delta \mathbf{v} = (4, 0) + (1, 3) = (5, 3) \). Since momentum is conserved and both pieces are of equal mass, the other piece moves at a velocity such that the total momentum vector remains \( (8, 0) \), so the velocity of the other piece is \( (3, -3) \).

Plot the Paths of the Two Pieces

Using their respective velocities:- For Piece 1 with \( \mathbf{v}_1 = (5, 3) \): \( x_1(t) = 16 + 5(t-4) \), \( y_1(t) = 8 + 3(t-4) - 0.5(t-4)^2 \).- For Piece 2 with \( \mathbf{v}_2 = (3, -3) \): \( x_2(t) = 16 + 3(t-4) \), \( y_2(t) = 8 - 3(t-4) - 0.5(t-4)^2 \).Plot both paths from \( t=4 \) to \( t=9 \) and mark the positions at each integer \( t \) value.

Verify Center of Mass (CM) Follows Original Path

Calculate the path of the CM:- \( x_{CM}(t) = \frac{x_1(t) + x_2(t)}{2} \).- \( y_{CM}(t) = \frac{y_1(t) + y_2(t)}{2} \).Ensure this CM path aligns with the original trajectory by verifying it traces the same parabolic curve: \( (4, 4) \) continuing with additional velocities added to the initial conditions.Verify by plotting and superimposing the CM path on the initial trajectory plot.

Key concepts

Conservation of Momentum

Momentum is a fundamental concept in physics that deals with the motion of objects. In any mechanical event, such as an explosion or collision, the total momentum of the system must remain constant if there are no external forces. This principle is known as the conservation of momentum.When the grenade in the exercise explodes, it splits into two pieces. One key point to remember is that the total momentum before the explosion equals the total momentum after. The grenade's velocity right before it explodes is given as \( \mathbf{v} = (4, 0) \). Since the pieces are of equal mass and the total momentum must stay at \( (8, 0) \), we calculate the other piece's velocity. We know one piece moves with additional velocity \( \Delta \mathbf{v} = (1, 3) \), making it \( (5, 3) \). Therefore, using conservation laws, the other piece must move at a velocity \( (3, -3) \) to ensure that the momentum vector sums up to the original total momentum. This ensures that the system's total momentum remains unchanged, highlighting the principle's effectiveness in predicting motion outcomes.

Trajectory Equation

The trajectory of a projectile is a path that it follows under the influence of gravity. It can be described using mathematical equations to predict its position over time.In this scenario, the grenade is initially thrown from the origin with a velocity \( \mathbf{v}_o = (4, 4) \). As it moves, its trajectory can be described by two separate equations for the horizontal and vertical positions:

• Horizontal position: \( x(t) = v_{ox} \cdot t = 4t \)
• Vertical position: \( y(t) = v_{oy} \cdot t - 0.5 \cdot g \cdot t^2 = 4t - 0.5t^2 \)

These equations split the motion into two independent components—horizontal and vertical. Gravity only affects the vertical component by continuously accelerating it downwards. By plotting these equations for the interval from \( t=0 \) to \( t=4 \), students can visualize how the grenade's path evolves over time, forming a parabola due to constant gravitational acceleration. Understanding these equations is key to connecting the mathematics of motion with real-world physics.

Center of Mass

The concept of the Center of Mass (CM) in a system depends on the distribution of mass and the relative positions of all parts. When dealing with multiple objects, the CM acts as a conceptual point where the mass of a system could be considered concentrated, and it moves as if all external forces were applied there.In this exercise, once the grenade splits into two equal pieces, we focus on tracing the path of their combined center of mass. The position of the CM at any point is given by the average of their individual positions:

• For finding \( x_{CM}(t) \), use \( x_{CM}(t) = \frac{x_1(t) + x_2(t)}{2} \)
• For finding \( y_{CM}(t) \), use \( y_{CM}(t) = \frac{y_1(t) + y_2(t)}{2} \)

Performing this calculation shows that the CM's path aligns perfectly with the grenade's original trajectory before it broke apart. This indicates that even after the explosion, the CM follows the same parabolic curve, highlighting an intrinsic property of projectile motion where no external forces interfere. Thus, the CM provides an insightful perspective on analyzing complex motions.