SATHEE: Motion in a Plane - Result Question 34

Motion in a Plane - Result Question 34

37. A particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:

(a) $2 m v$

(b) $m v / \sqrt{2}$

(d) zero

[2008]

Show Answer

Answer:

Correct Answer: 37. (c)

Solution:

Clearly, change in momentum along $x$ -axis $= m v \cos θ - m v \cos θ = 0$

Momentum changed only in vertical direction or $y$ -axis.

$\begin{aligned} So, & Δ P = Δ P_{vertical} \\ \Rightarrow & P_{final} = P_{initial} \\ = m v \sin θ - (- m v \sin θ) \\ = 2 m v \sin θ = 2 m v \times \sin 45^{\circ} \\ = 2 m v \times \frac{1}{\sqrt{2}} = \sqrt{2} m v \end{aligned}$

Hence, resultant change in momentum $= \sqrt{2} m v$

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Motion in a Plane - Result Question 34

37. A particle of mass m is projected with velocity v making an angle of 45∘ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:

Answer:

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37. A particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be: