Train With The Constant Acceleration

Classical Mechanics Level 4

A train is moving along a straight line with constant acceleration a a . A body standing in the train throws a ball forward with a speed of 10 m/s 10\text{ m/s} at an angle of 6 0 ∘ 60^\circ to the horizontal. The body has to move forward by 1.15 m 1.15\text{ m} inside to catch the ball back at the initial height.Find the acceleration of train ( in m/s 2 \text{m/s}^2 ).

Details and Assumptions :
Take g = 10 m/s 2 g =10 \text{ m/s}^2 and use the estimation 3 = 1.73 \sqrt{3} =1.73 .


The answer is 5.

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Ashish Menon
Oct 14, 2016

The body shows projectile motion, so its x-component of velocity is v cos ⁡ θ v\cos\theta = 10 cos ⁡ 60 ° = 5 10\cos{60}^° = 5 .
Now, the time of flight of a projectile is 2 v sin ⁡ θ g = 2 × 10 × 3 2 × 10 = 2 \dfrac{2v\sin\theta}{g} = \dfrac{2×10×\sqrt{3}}{2×10} = \sqrt{2} .

Applying formula S = u t + 1 2 a t 2 1.15 = 5 × 3 − 1 2 a × ( 3 ) 2 1.15 = 5 × 1.73 − 3 2 a a = 5 S = ut + \dfrac{1}{2}at^2\\ \\ 1.15 = 5×\sqrt{3} - \dfrac{1}{2}a×{\left(\sqrt{3}\right)}^2\\ \\ 1.15 = 5×1.73 - \dfrac{3}{2}a\\ \\ a = \boxed{5}

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Md Zuhair
Sep 26, 2016

How to do @Ashish Siva

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?? i got the answer 5

Ashish Menon - 4 years, 8 months ago

The body shows projectile motion, so its x-component of velocity is v cos ⁡ θ v\cos\theta = 10 cos ⁡ 60 ° = 5 10\cos{60}^° = 5 .
Now, the time of flight of a projectile is 2 v sin ⁡ θ g = 2 × 10 × 3 2 × 10 = 2 \dfrac{2v\sin\theta}{g} = \dfrac{2×10×\sqrt{3}}{2×10} = \sqrt{2} .

Applying formula S = u t + 1 2 a t 2 1.15 = 5 × 3 − 1 2 a × ( 3 ) 2 1.15 = 5 × 1.73 − 3 2 a a = 5 S = ut + \dfrac{1}{2}at^2\\ \\ 1.15 = 5×\sqrt{3} - \dfrac{1}{2}a×{\left(\sqrt{3}\right)}^2\\ \\ 1.15 = 5×1.73 - \dfrac{3}{2}a\\ \\ a = \boxed{5}

Ashish Menon - 4 years, 8 months ago

Are you sure of the answer? An acceleration of 0.2 does not seem to satisfy :/. Let me check once more.

Ashish Menon - 4 years, 8 months ago

Exactly,the answer is 5 @Ashish Siva the projectile is sort of a returning boomerang here,as at t=1 the horizontal component reduces to zero and till t=√3 it traces path towards the point of projection!! So the acceleration is 5ms^-2!!

Chirag Shyamsundar - 4 years, 8 months ago

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I also think so. Thanks. I will tell them to change the answer.

Md Zuhair - 4 years, 8 months ago

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Thanks for sharing this nice question :)

Ashish Menon - 4 years, 8 months ago

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@Ashish Menon – Welcome my friend...

Md Zuhair - 4 years, 8 months ago

@Calvin Lin . Please change the answer is 5.

Md Zuhair - 4 years, 8 months ago

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Thanks. I have updated the answer to 5.

Calvin Lin Staff - 4 years, 8 months ago

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