Question

A bullet of mass 10 g moving horizontally at a speed of 50√7 m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure. The bullet remains inside the block and the system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?
Figure

Answer 1

Given:
Mass of block = 490 gm
Initial speed of the block = 0
Mass of bullet = 10 gm
Initial speed of the bullet, v_{1 =} $50\sqrt{7}\mathrm{m}/\mathrm{s}$

As the bullet gets embedded inside the block, this is an example of a perfectly inelastic collision.

Let the final velocity of the system (block and bullet) be V_A.

Using the law of conservation of linear momentum, we get:
$$m₁v₁ + m₂ × 0 = (m₁ + m₂)v_A
⇒10 × 10 - 3 × 507 + 490 × 10 - 3 × 0 = (490 + 10) × 10 - 3 × v_A
∴ v_A=$\sqrt{7}$ m/s

When the block loses contact at D, the component mg acts on it.
Let the velocity at D be v_B.



$$\begin{gathered} \frac{m(v_{\mathrm{B}}{)}^2}{r}=mg\sin \theta \\ (v_{\mathrm{B}}{)}^2=gr\sin \theta ...(1) \end{gathered}$$
Using work energy theorem
$$\begin{gathered} \left(\frac{1}{2}\right)m{v}_{\mathrm{B}}^2-\left(\frac{1}{2}\right)m{v}_{\mathrm{A}}^2=-mg(0.2+0.2\sin \theta ) \\ \left(\frac{1}{2}\right)gr\sin \theta -\left(\frac{1}{2}\right){\left(\sqrt{7}\right)}^2=g(0.2+0.2\sin \theta ) \end{gathered}$$

$$\begin{gathered} 3.5-\frac{1}{2}+9.8+0.2+\sin \theta =9.8+0.2(1+\sin \theta ) \\ 3.5-0.98\sin \theta =1.96+1.96\sin \theta \\ \sin \theta =\frac{1}{2} \\ \Rightarrow \theta ={30}^{\mathrm{o}} \end{gathered}$$

∴ Angle of projection = 90° − 30° = 60°

Time of reaching the ground,

$$\begin{gathered} \Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times (0.2+0.2\sin {30}^{\mathrm{o}})}{9.8}} \\ \Rightarrow t=0.247\mathrm{s} \end{gathered}$$

Distance travelled in the horizontal direction is given by,
S = v cos θ × t
$$
S = gr sin θ cos θ×t
S = 9.8 × 2 × 1232 × 0.247
S = 0.196 m

Total distance = (0.2 − 0.2 cos 30° + 0.196)
= 0.22 m