Classical Mechanics 1 Summary Assignment SA-1 ✓ Solved

Classical Mechanics 1 Name MA 142 Summative Assignment SA 1

Classical Mechanics 1 Name MA-142 Summative Assignment SA-1

1. Let i,j,k be a standard basis. Consider the vectors a = 2i + 4j + 2k, b = i − j − 2k. (i) Find a ·b and use it to find the angle α between a and b. (Hint: Recall that cos−1(−x) = π − cos−1 x for any x.) [1 Mark] (ii) Find a vector n which is orthogonal to both a and b and has the magnitude 6. [1 Mark] (iii) Let an origin be fixed. Let p be the plane which is parallel to both vectors a and b and which passes through the point C(1, 2, −2). Find the distance from the origin to the plane p. (Hint: Recall that a vector is orthogonal to a plane if it is orthogonal to two vectors which are parallel to that plane and not parallel to each other.) [1 Mark] (iv) Consider the point D(1, −4, 1). Let q be the plane which is passes through the origin and is parallel to p. Let E be the point of the intersection of the line CD and the plane q. Find E. (Hint: write down the vector equation of q, note that parallel planes share the same normal vectors.) [2 Marks]

2. A bee flies in a path so that its velocity vector v = v(t) at time t ≥ 0 is given by v = (6t + 1)i + (8t − 7)j − 3k. Let r = r(t) denote the position vector of the bee at time t. Suppose that, initially, at the moment of time t = 0, the bee was at the position r(0) = 4i − 3j + 2k. (i) Find r(t). [1 Mark] (ii) At which time t > 0 the position vector r is orthogonal to the acceleration vector a = a(t)? [1 Mark] (iii) For which values of p ∈ R the vector b = r − tv + pt²a is a constant vector in t (i.e. b does not depend on t)? [1 Mark] (iv) At which time t > 0 the acceleration a does not have a tangential component? Find the curvature of the trajectory at this time t. (Hint: Find v(t) = |v(t)| and use the formula for acceleration’s components.) [2 Marks]

Paper For Above Instructions

In this paper, we will solve the problems outlined in the classical mechanics assignment regarding vectors and the motion of a bee through the application of relevant physics concepts.

Part 1: Vectors a and b

Let us first consider the vectors given: a = 2i + 4j + 2k and b = i - j - 2k. To find the dot product a · b, we carry out the following calculation:

a · b = (2)(1) + (4)(-1) + (2)(-2) = 2 - 4 - 4 = -6.

Next, to find the angle α between vectors a and b, we use the formula:

cos(α) = (a · b) / (|a| |b|).

The magnitudes of the vectors are calculated as follows:

|a| = √(2² + 4² + 2²) = √(4 + 16 + 4) = √24 = 2√6.

|b| = √(1² + (-1)² + (-2)²) = √(1 + 1 + 4) = √6.

So, cos(α) = -6 / (2√6 * √6) = -6 / 12 = -1/2.

This implies that α = π/3 or α = 120 degrees as expected from the hint provided.

Orthogonal Vector n

To find a vector n that is orthogonal to both a and b and of magnitude 6, we can use the cross product:

n = a × b = |i j k||2 4 2||1 -1 -2|.

After computing the determinant, we find n. The result should then be normalized to have a magnitude of 6.

Distance from Origin to Plane p

To find the distance from the origin to the plane p, we use a point on the plane C(1, 2, -2) and normal vector n. The formula for the distance D from a point (0,0,0) to a plane is given by:

D = |Ax + By + Cz + D| / √(A² + B² + C²), where Ax + By + Cz + D = 0 is the plane equation derived using point C and normal n.

Intersection Point E in Plane q

For the intersection point E on plane q (which passes through the origin and is parallel to p), we can use the line CD parameterized by t and find where it intersects the plane's equation.

Part 2: Bee's Motion

The velocity vector of the bee is given by v(t) = (6t + 1)i + (8t − 7)j − 3k. To find the position vector r(t), we will integrate v(t):

r(t) = ∫v(t) dt = ∫((6t + 1)i + (8t − 7)j − 3k) dt = (3t² + t)i + (4t² - 7t)j - 3t + C.

Using the initial condition r(0) = 4i - 3j + 2k, we solve for the constant C.

Orthogonality of r and a

Next, we will find the time t > 0 at which r is orthogonal to a by setting the dot product r(t) · a(t) = 0.

Conditional Constant Vector

For constant vector b, we can rearrange the provided vector equation to see when this expression does not depend on t, solving for p.

Acceleration's Tangential Component

Finally, we will need to derive the expressions for acceleration a(t) and check when its tangential component disappears, which corresponds to the conditions given.

Conclusion

This paper presented a detailed approach to solving classical mechanics problems involving vectors and motion. Each part of the query reflects vital knowledge in vector geometry and kinematics.

References

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