Menu Close

A-50g-golf-ball-is-struck-with-a-club-moving-with-a-velocity-of-22m-s-after-it-moves-4cm-and-ball-accelerates-with-a-velocity-of-44m-s-estimate-the-average-force-exerted-by-the-club-on-the-ball-




Question Number 114114 by aurpeyz last updated on 17/Sep/20
  A 50g golf ball is struck with a club moving  with a velocity of 22m/s after it moves 4cm and  ball accelerates with a velocity of 44m/s.  estimate the average force exerted by the club  on the ball.
$$ \\ $$$$\mathrm{A}\:\mathrm{50g}\:\mathrm{golf}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{struck}\:\mathrm{with}\:\mathrm{a}\:\mathrm{club}\:\mathrm{moving} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{22m}/\mathrm{s}\:\mathrm{after}\:\mathrm{it}\:\mathrm{moves}\:\mathrm{4cm}\:\mathrm{and} \\ $$$$\mathrm{ball}\:\mathrm{accelerates}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{44m}/\mathrm{s}. \\ $$$$\mathrm{estimate}\:\mathrm{the}\:\mathrm{average}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{by}\:\mathrm{the}\:\mathrm{club} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{ball}. \\ $$
Commented by aurpeyz last updated on 17/Sep/20
pls help
$${pls}\:{help} \\ $$
Answered by Ar Brandon last updated on 17/Sep/20
m_b =0.05Kg, u_b =0m/s, v_b =44m/s, m_c =?, u_c =22m/s  v_c =0m/s. Momentum is conserved  ⇒m_b u_b +m_c u_c =m_b v_b +m_c v_c ⇒22m_c =44×0.05  ⇒m_c =0.1Kg  For club;  v=0m/s, a=?, u=22m/s, s=4cm=0.04m  Using v^2 =u^2 +2as, 0=22^2 +2a×0.04⇒a=−6050m/s^2   ⇒F=ma=0.1×6050=605 N
$$\mathrm{m}_{\mathrm{b}} =\mathrm{0}.\mathrm{05Kg},\:\mathrm{u}_{\mathrm{b}} =\mathrm{0m}/\mathrm{s},\:\mathrm{v}_{\mathrm{b}} =\mathrm{44m}/\mathrm{s},\:\mathrm{m}_{\mathrm{c}} =?,\:\mathrm{u}_{\mathrm{c}} =\mathrm{22m}/\mathrm{s} \\ $$$$\mathrm{v}_{\mathrm{c}} =\mathrm{0m}/\mathrm{s}.\:\mathrm{Momentum}\:\mathrm{is}\:\mathrm{conserved} \\ $$$$\Rightarrow\mathrm{m}_{\mathrm{b}} \mathrm{u}_{\mathrm{b}} +\mathrm{m}_{\mathrm{c}} \mathrm{u}_{\mathrm{c}} =\mathrm{m}_{\mathrm{b}} \mathrm{v}_{\mathrm{b}} +\mathrm{m}_{\mathrm{c}} \mathrm{v}_{\mathrm{c}} \Rightarrow\mathrm{22m}_{\mathrm{c}} =\mathrm{44}×\mathrm{0}.\mathrm{05} \\ $$$$\Rightarrow\mathrm{m}_{\mathrm{c}} =\mathrm{0}.\mathrm{1Kg} \\ $$$$\mathrm{For}\:\mathrm{club}; \\ $$$$\mathrm{v}=\mathrm{0m}/\mathrm{s},\:\mathrm{a}=?,\:\mathrm{u}=\mathrm{22m}/\mathrm{s},\:\mathrm{s}=\mathrm{4cm}=\mathrm{0}.\mathrm{04m} \\ $$$$\mathrm{Using}\:\mathrm{v}^{\mathrm{2}} =\mathrm{u}^{\mathrm{2}} +\mathrm{2as},\:\mathrm{0}=\mathrm{22}^{\mathrm{2}} +\mathrm{2a}×\mathrm{0}.\mathrm{04}\Rightarrow\mathrm{a}=−\mathrm{6050m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{F}=\mathrm{ma}=\mathrm{0}.\mathrm{1}×\mathrm{6050}=\mathrm{605}\:\mathrm{N} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *