A bar of mass M = 1.00 kg and length L = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. [2023]

Question

A bar of mass M = 1.00 kg and length L = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m = 0.10 kg is moving on the same horizontal surface with $5.00 {m s}^{- 1}$ speed on a path perpendicular to the bar. It hits the bar at a distance L/2 from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity $ω$ . Which of the following statement is correct? [2023]

Smart Achievers · Accepted Answer

(1)

About the hinge applying angular momentum conservation

$m v \frac{L}{2} + 0 = - m v \frac{L}{2} + \frac{M L^{2}}{3} ω . . . (i)$

After elastic collision,

$e = 1 = \frac{ω \frac{L}{2} + V}{u} \Rightarrow u = ω \frac{L}{2} + V . . . (i i)$

$Putting, m = 0.1 kg, M = 1 kg, L = 0.20 m and solving eq (i) & (ii) we get$

$ω \approx 6.98 {rad s}^{- 1} and v = 4.30 {m s}^{- 1}$

Solution

1 $ω = 6.98 {rad s}^{- 1}$ and $v = 4.30 {m s}^{- 1}$

2 $ω = 3.75 {rad s}^{- 1}$ and $v = 4.30 {m s}^{- 1}$

3 $ω = 3.75 {rad s}^{- 1}$ and $v = 10.0 {m s}^{- 1}$

4 $ω = 6.80 {rad s}^{- 1}$ and $v = 4.10 {m s}^{- 1}$

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Solution

1 ω=6.98 rad s-1 and v=4.30 m s-1

2 ω=3.75 rad s-1 and v=4.30 m s-1

3 ω=3.75 rad s-1 and v=10.0 m s-1

4 ω=6.80 rad s-1 and v=4.10 m s-1

Ans. (1) About the hinge applying angular momentum conservation mvL2+0=-mvL2+ML23ω ...(i) After elastic collision, e=1=ωL2+Vu⇒u=ωL2+V ...(ii) Putting, m=0.1 kg, M=1 kg, L=0.20 m and solving eq (i) & (ii) we get ω≈6.98 rad s-1 and v=4.30 m s-1

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1 $ω = 6.98 {rad s}^{- 1}$ and $v = 4.30 {m s}^{- 1}$

2 $ω = 3.75 {rad s}^{- 1}$ and $v = 4.30 {m s}^{- 1}$

3 $ω = 3.75 {rad s}^{- 1}$ and $v = 10.0 {m s}^{- 1}$

4 $ω = 6.80 {rad s}^{- 1}$ and $v = 4.10 {m s}^{- 1}$