Question

Figure shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and$$\begin{gathered} v_m=u=a_{2}t \\ =\frac{mg\sin \alpha \cos \alpha }{\mathrm{M}+m}\sqrt{\frac{2h}{a\sin \alpha }}={\left[\frac{2m^2{g}^{2}h{\sin }^2\alpha c{o}^2\alpha }{(\mathrm{M}+m{)}^{2}a\sin \alpha }\right]}^{1/2} \end{gathered}$$ the system is left to itself. Assume that all the surface are frictionless. (a) Find the speed of the larger block when the smaller block is sliding on the vertical part. (b) Find the speed of the smaller mass when it breaks off the larger mass at height h. (c) Find the maximum height (from the ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distance traversed by the bigger block during the time when the smaller block was in its flight under gravity.
Figure

Answer 1

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According to the question, mass m is given with a speed v over the larger mass M.

(a) When the smaller block travels on the vertical part, let the velocity of the bigger block be v₁, towards left.

From the law of conservation of momentum (in the horizontal direction), we get:
$$\begin{gathered} mv=(\mathrm{M}+m)v_1 \\ \Rightarrow v_1=\frac{mv}{M+m} \end{gathered}$$

(b) When the smaller block breaks off, let its resultant velocity be v₂.

Using the law a of conservation of energy, we get:

$$\begin{gathered} \left(\frac{1}{2}\right)m{v}^2=\left(\frac{1}{2}\right)\mathrm{M}{v}_1^2+\left(\frac{1}{2}\right)m{v}_2^2+\mathrm{mgh}\text{ \\ ⇒v22=v2-Mmv12-2gh...(i) \\ ⇒v22=v21-Mm.m2(M+m)2-2gh \\ ⇒v2=m2+Mm+m2v2(M+m)2-2gh1/2} \end{gathered}$$

(c) Vertical component of the velocity v₂ of mass m is given by,

$$\begin{gathered} {v_y}^2={v_2}^2-{v_1}^2\text{ \\ =M2+Mm+m2M+m2v2-2gh-m2v2M+m2 \\ ∴v1=mvM+m \\ ⇒vy2=M2+Mm+m2-m2M+m2v2-2gh \\ ⇒vy2=Mv2M+m-2gh...(ii)} \end{gathered}$$

To determine the maximum height (from the ground),
Let us assume that the body rises to a height h₁ over and above h.

$$\begin{gathered} \mathrm{Now},\left(\frac{1}{2}\right)m{v}_y^2=mg{h}_1\text{ \\ ⇒h1=vy22g...(iii) \\ ∴Totalheight,H=h+h1 \\ ⇒H=h+vy22g \\ ⇒H=h+Mv2(M+m)2g-h} \end{gathered}$$

$\Rightarrow H=\frac{M{v}^2}{(M+m)2g}$

(d) Because the smaller mass also has a horizontal component of velocity V₁ at the time it breaks off from M (that has a velocity v₁), the block m will again land on the block M.

The time of flight of block m after it breaks off is calculated as:

During the upward motion (BC),
$$\begin{gathered} 0=v_y-gt\text{ \\ ⇒t1=vyg=1gMv2(M+m)-2gh...(4) \\ } \end{gathered}$$

Thus, the time for which the smaller block is in flight is given by,
$$\begin{gathered} T=2t_1 \\ =\frac{2}{g}\left[\frac{M{v}^2-2(M+m)gh}{(M+m)}\right] \end{gathered}$$

The distance travelled by the bigger block during this time is,
$S=v_{1}T$