Collisions in Two Dimensions — Question 9

Section 9-10 · Physics pages 1348–1349

Question 9 Two upward departures contradict zero transverse input A 1 kg projectile moving along +x hits a stationary 2 kg target. Both final speeds are strictly positive. A report places both outgoing directions between 10◦and 30◦above +x. The system is said to receive no external impulse.⃗ v1⃗ v2 Both reported transverse components have the same positive sign. (a) Test whether any speeds can satisfy momentum conservation. (b) If the speeds are measured as 3 and 2 m/s, bound the necessary external y impulse.
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Question 9 - Solution Two upward departures contradict zero transverse input⃗ v1⃗ v2 (a) Derivation and calculation The initial transverse momentum vanishes, but the reported final value is Pfy = m1v1 sin θ1 + m2v2 sin θ2. With positive masses and speeds and 10◦≤θi ≤30◦, Pfy > 0 = Piy ⇒zero external impulse is impossible . (b) Result and interpretation For the measured speeds, Jy = 3 sin θ1 + 4 sin θ2. Sine is increasing throughout the allowed angular range, so 7 sin 10◦≤Jy ≤7 sin 30◦, 1.216 ≤Jy ≤3.50 N s . Physical check No energy assumption can repair a transverse momentum contradiction.

Question 9

Two upward departures contradict zero transverse input

A 1kg1\,\mathrm{kg} projectile moving along +x+x hits a stationary 2kg2\,\mathrm{kg} target. Both final speeds are strictly positive. A report places both outgoing directions between 10∘10^\circ and 30∘30^\circ above +x+x. The system is said to receive no external impulse.

Diagram shown in the original SVG worksheet.

  1. Test whether any speeds can satisfy momentum conservation.

  2. If the speeds are measured as 33 and 2m/s2\,\mathrm{m/s}, bound the necessary external yy impulse.

Question 9 - Solution

Two upward departures contradict zero transverse input

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The initial transverse momentum vanishes, but the reported final value is Pfy=m1v1sin⁡θ1+m2v2sin⁡θ2.P_{fy}=m_1v_1\sin\theta_1+m_2v_2\sin\theta_2. With positive masses and speeds and 10∘≤θi≤30∘10^\circ\leq\theta_i\leq30^\circ, Pfy>0=Piy⇒zero external impulse is impossible.P_{fy}>0=P_{iy}\Rightarrow\boxed{\text{zero external impulse is impossible}}.

(b) Result and interpretation

For the measured speeds, Jy=3sin⁡θ1+4sin⁡θ2.J_y=3\sin\theta_1+4\sin\theta_2. Sine is increasing throughout the allowed angular range, so 7sin⁡10∘≤Jy≤7sin⁡30∘,1.216≤Jy≤3.50Ns.\boxed{7\sin10^\circ\leq J_y\leq7\sin30^\circ},\qquad\boxed{1.216\leq J_y\leq3.50\ \mathrm{N\,s}}.

Physical check

No energy assumption can repair a transverse momentum contradiction.