Question

A shell of mass 1kg is projected with velocity 20 m/s at an angle 60${}^{\circ }$with horizontal. It collides inelastically with a ball of mass 1kg which is suspended through a thread of length 1m. The other end of the thread is attached to the ceiling of a trolley of mass$\frac{4}{3}$ kg as shown in figure. Initially the trolley is stationary and it is free to move along horizontal rails without any friction. If the maximum reflection of the thread with vertical is 12x ${}^{\circ }$ for which, string does not slack. Then x is : $(Takeg=10m/s^2)$

Answer 1

H=$\frac{u^{2}si{n}^2\theta }{2g}=\frac{{\left(20\right)}^2\times \frac{3}{4}}{2\times 10}$

$=15m$

i.e., the shell strikes the ball at highest point of its trajectory.

Velocity of (ball + shell) just after collision.

v=$\frac{ucos{60}^{\circ }}{2}$ (from conservation of linear momentum)

$=\frac{20}{2\times 2}$=5m/s

At highest point combined mass is at rest relative to the trolley.

Let v be the velocity of trolley at this instant. From conservation of linear momentum we have,

$2\times 5=(2+\frac{3}{4})v$

$v=3m/s$

From conservation of energy, we have

$\frac{1}{2}\times 2\times (5{)}^2-\frac{1}{2}(2+\frac{4}{3})(3{)}^2=2\times 10(1-cos\theta )$

solving we get,

$cos\theta =\frac{1}{2}$

$\therefore \theta ={60}^{\circ }$