Topic : Collisions Concepts : Conservation of linear momentum , Impulse-momentum theorem , Relative velocity
Solution :
Since there are no external forces, we have conservation of linear momentum. Initially, the astronaut is at rest so the initial momentum
After the launcher is fired,
p f = m ast v 1 − m ball v 2 p_f=m_{\text{ast}}v_1-m_{\text{ball}}v_2 Since p i = p f p_i=p_f
m ast v 1 = m ball v 2 m_{\text{ast}}v_1=m_{\text{ball}}v_2 We are given the relative velocity v 1 + v 2 = v r v_1+v_2=v_r
m ast v 1 = m ball ( v r − v 1 ) m_{\text{ast}}v_1=m_{\text{ball}}(v_r-v_1) Solving for v 1 v_1
m ast v 1 + m ball v 1 = m ball v r m_{\text{ast}}v_1+m_{\text{ball}}v_1=m_{\text{ball}}v_r v 1 = m ball v r m ast + m ball v_1=\frac{m_{\text{ball}}v_r}{m_{\text{ast}}+m_{\text{ball}}} Thus, the impulse delivered to the astronaut is
J = m ast v 1 = m ast m ball v r m ast + m ball = ( 100 kg ) ( 10 kg ) ( 50 m / s ) 100 kg + 10 kg = 455 N s J=m_{\text{ast}}v_1=\frac{m_{\text{ast}}m_{\text{ball}}v_r}{m_{\text{ast}}+m_{\text{ball}}}=\frac{(100\,\mathrm{kg})(10\,\mathrm{kg})(50\,\mathrm{m/s})}{100\,\mathrm{kg}+10\,\mathrm{kg}}=455\,\mathrm{N\,s} so the answer is A .
Topic : Collisions Concepts : Impulse-momentum theorem , Vectors
Solution :
The astronaut has some initial momentum p → i = m ast v → \vec{p}_i=m_{\text{ast}}\vec{v} J = 455 N s J=455\,\mathrm{N\,s}
p → f = p → i + J → \vec{p}_f=\vec{p}_i+\vec{J} the tip of the final momentum vector can be anywhere on the sphere reachable by J → \vec{J} θ \theta p → i \vec{p}_i p → f \vec{p}_f p → f \vec{p}_f
sin θ = J p i \sin\theta=\frac{J}{p_i} θ = arcsin ( J m ast v ) = arcsin ( 455 100 ⋅ 10 ) = 27 ∘ \theta=\arcsin\left(\frac{J}{m_{\text{ast}}v}\right)=\arcsin\left(\frac{455}{100 \cdot 10}\right)=27^{\circ} so the answer is C .
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