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Question Number 18142 by Tinkutara last updated on 15/Jul/17
An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0, the plank starts moving along the x- direction with an acceleration 1.5 ms^(−2) . At the same instant a stone is projected from the origin with a velocity u^→ as shown. A stationary person on the ground observe the stone hitting the object during its downward motion at an angle of 45° with the horizontal. Take g = 10 m/s^2 and consider all motions in the x-y plane. 1. The time after which the stone hits the object is 2. The initial velocity (u^→ ) of the particle is
$$\mathrm{An}\:\mathrm{object}\:{A}\:\mathrm{is}\:\mathrm{kept}\:\mathrm{fixed}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point} \\ $$$${x}\:=\:\mathrm{3}\:\mathrm{m}\:\mathrm{and}\:{y}\:=\:\mathrm{1}.\mathrm{25}\:\mathrm{m}\:\mathrm{on}\:\mathrm{a}\:\mathrm{plank}\:{P} \\ $$$$\mathrm{raised}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{At}\:\mathrm{time}\:{t}\:=\:\mathrm{0}, \\ $$$$\mathrm{the}\:\mathrm{plank}\:\mathrm{starts}\:\mathrm{moving}\:\mathrm{along}\:\mathrm{the}\:{x}- \\ $$$$\mathrm{direction}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}\:\mathrm{1}.\mathrm{5}\:\mathrm{ms}^{−\mathrm{2}} . \\ $$$$\mathrm{At}\:\mathrm{the}\:\mathrm{same}\:\mathrm{instant}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\overset{\rightarrow} {{u}}\:\mathrm{as} \\ $$$$\mathrm{shown}.\:\mathrm{A}\:\mathrm{stationary}\:\mathrm{person}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{observe}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{hitting}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{during}\:\mathrm{its}\:\mathrm{downward}\:\mathrm{motion}\:\mathrm{at} \\ $$$$\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{45}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}. \\ $$$$\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{consider}\:\mathrm{all} \\ $$$$\mathrm{motions}\:\mathrm{in}\:\mathrm{the}\:{x}-{y}\:\mathrm{plane}. \\ $$$$\mathrm{1}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{after}\:\mathrm{which}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{object}\:\mathrm{is} \\ $$$$\mathrm{2}.\:\mathrm{The}\:\mathrm{initial}\:\mathrm{velocity}\:\left(\overset{\rightarrow} {{u}}\right)\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 15/Jul/17
Answered by ajfour last updated on 16/Jul/17
let initial position of object be P(a,b) ⇒ a=3m , b=1.25m let stone hits object at x=d, y=b. let time taken =t_0 In this time object moves from x=a to x=d with acceleration A=1.5m/s^2 . So d=a+(1/2)At_0 ^2 ....(i) As stone hits object while descending at α=45° , we observe v_y =−u_x , v_x =u_x ....(ii) d=u_x t_0 ...(iii) from (i) and (iii): a+((At_0 ^2 )/2)=u_x t_0 ...(iv) Also b=u_y t_0 −((gt_0 ^2 )/2) ....(v) And [ v_y =−u_x =u_y −gt_0 ] ×t_0 or −u_x t_0 =u_y t_0 −gt_0 ^2 ...(vi) (v)−(vi) gives b+u_x t_0 =((gt_0 ^2 )/2) . And from eq.(iv) b+(a+((At_0 ^2 )/2))=((gt_0 ^2 )/2) substituting values (5/4)+3+(3/4)t_0 ^2 =5t_0 ^2 17t_0 ^2 =17 ⇒ t_0 ^2 =1 ⇒ t_0 =1s so u_x =(a/t_0 )+((At_0 )/2) = 3+(3/4)= ((15)/4)m/s^2 u_y =(b/t_0 )+((gt_0 )/2) = (5/4)+5 =((25)/4)m/s^2 u^→ =(5/4)(3i^� +5j^� )m/s .
$$\mathrm{let}\:\mathrm{initial}\:\mathrm{position}\:\mathrm{of}\:\mathrm{object}\:\mathrm{be} \\ $$$$\:\mathrm{P}\left(\mathrm{a},\mathrm{b}\right)\:\:\:\Rightarrow\:\:\mathrm{a}=\mathrm{3m}\:\:,\:\mathrm{b}=\mathrm{1}.\mathrm{25m} \\ $$$$\mathrm{let}\:\mathrm{stone}\:\mathrm{hits}\:\mathrm{object}\:\mathrm{at}\:\mathrm{x}=\mathrm{d},\:\mathrm{y}=\mathrm{b}. \\ $$$$\:\mathrm{let}\:\mathrm{time}\:\mathrm{taken}\:=\mathrm{t}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\mathrm{In}\:\mathrm{this}\:\mathrm{time}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{from} \\ $$$$\:\:\:\mathrm{x}=\mathrm{a}\:\:\mathrm{to}\:\:\:\mathrm{x}=\mathrm{d}\:\:\:\mathrm{with}\:\mathrm{acceleration} \\ $$$$\:\:\:\mathrm{A}=\mathrm{1}.\mathrm{5m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{So} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{d}=\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{At}_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\:\:\:\:….\left(\mathrm{i}\right) \\ $$$$\:\:\:\mathrm{As}\:\mathrm{stone}\:\mathrm{hits}\:\mathrm{object}\:\mathrm{while} \\ $$$$\mathrm{descending}\:\mathrm{at}\:\alpha=\mathrm{45}°\:,\:\mathrm{we}\:\mathrm{observe} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{v}_{\mathrm{y}} =−\mathrm{u}_{\mathrm{x}} \:,\:\:\mathrm{v}_{\mathrm{x}} =\mathrm{u}_{\mathrm{x}} \:\:\:\:….\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{d}=\mathrm{u}_{\mathrm{x}} \mathrm{t}_{\mathrm{0}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left(\mathrm{iii}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{iii}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\frac{\mathrm{At}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}=\mathrm{u}_{\mathrm{x}} \mathrm{t}_{\mathrm{0}} \:\:\:\:\:\:\:\:…\left(\mathrm{iv}\right) \\ $$$$\:\mathrm{Also}\:\:\:\:\mathrm{b}=\mathrm{u}_{\mathrm{y}} \mathrm{t}_{\mathrm{0}} −\frac{\mathrm{gt}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:….\left(\mathrm{v}\right) \\ $$$$\mathrm{And}\:\:\:\:\left[\:\mathrm{v}_{\mathrm{y}} =−\mathrm{u}_{\mathrm{x}} =\mathrm{u}_{\mathrm{y}} −\mathrm{gt}_{\mathrm{0}} \:\:\:\right]\:\:×\mathrm{t}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\mathrm{or}\:\:\:\:−\mathrm{u}_{\mathrm{x}} \mathrm{t}_{\mathrm{0}} =\mathrm{u}_{\mathrm{y}} \mathrm{t}_{\mathrm{0}} −\mathrm{gt}_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\:…\left(\mathrm{vi}\right) \\ $$$$\left(\mathrm{v}\right)−\left(\mathrm{vi}\right)\:\mathrm{gives} \\ $$$$\:\:\:\:\mathrm{b}+\mathrm{u}_{\mathrm{x}} \mathrm{t}_{\mathrm{0}} =\frac{\mathrm{gt}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}\:\:.\:\mathrm{And}\:\mathrm{from}\:\mathrm{eq}.\left(\mathrm{iv}\right) \\ $$$$\:\:\mathrm{b}+\left(\mathrm{a}+\frac{\mathrm{At}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{gt}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{substituting}\:\mathrm{values} \\ $$$$\:\:\:\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{3}+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{t}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{5t}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{17t}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{17} \\ $$$$\Rightarrow\:\:\:\mathrm{t}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{1}\:\:\:\:\:\: \\ $$$$\Rightarrow\:\:\mathrm{t}_{\mathrm{0}} =\mathrm{1s} \\ $$$$\mathrm{so}\:\:\mathrm{u}_{\boldsymbol{\mathrm{x}}} =\frac{\mathrm{a}}{\mathrm{t}_{\mathrm{0}} }+\frac{\mathrm{At}_{\mathrm{0}} }{\mathrm{2}}\:=\:\mathrm{3}+\frac{\mathrm{3}}{\mathrm{4}}=\:\frac{\mathrm{15}}{\mathrm{4}}\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{u}_{\mathrm{y}} =\frac{\mathrm{b}}{\mathrm{t}_{\mathrm{0}} }+\frac{\mathrm{gt}_{\mathrm{0}} }{\mathrm{2}}\:=\:\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{5}\:=\frac{\mathrm{25}}{\mathrm{4}}\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{u}}=\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{3}\hat {\mathrm{i}}+\mathrm{5}\hat {\mathrm{j}}\right)\mathrm{m}/\mathrm{s}\:\:.\:\:\:\:\: \\ $$
Commented by ajfour last updated on 16/Jul/17
view again, it is shorter now.
$$\mathrm{view}\:\mathrm{again},\:\mathrm{it}\:\mathrm{is}\:\mathrm{shorter}\:\mathrm{now}. \\ $$
Commented by Tinkutara last updated on 16/Jul/17
Now it is better.
$$\mathrm{Now}\:\mathrm{it}\:\mathrm{is}\:\mathrm{better}. \\ $$
Commented by ajfour last updated on 16/Jul/17
Is my answer correct?
$$\mathrm{Is}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{correct}? \\ $$
Commented by Tinkutara last updated on 16/Jul/17
Yes, Thanks Sir!
$$\mathrm{Yes},\:\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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