Question

A block of mass 9 kg is lying on a rough horizontal surface. A particle of mass 1 kg strikes the block with a speed of 50 m/s at an angle of 37 with the vertical and sticks to it. The frictional coefficient between the ground and the block is $\mu$ =0.4. Find the velocity of the combined system just after the collision assuming limiting friction coefficient between the block and the ground after the collision?

Answer 1

Case $\left(1\right)$, if $\mu =0$

Momentum changes by block in collision

$f_2△t=9v⟶\left(1\right)$

By momentum coservation in $\alpha$-direction,

$1\times 50\sin 37°=(9+1)vv=3m/s$

Equation $\left(1\right)$ becomes,

$f_2△t=27$

Momentum changed in $y$-direction by ball,

$f_1△t=50\cos 37°=40$ms.

When $△t$ is very samll collision time,

if $\mu \ne 0=0.4$

Frictional force=$(F_1+10g)\mu$

In $x$-direction momentum change by block=$9v^{\prime }$

${(f}_2-fs)△t=9v^{\prime }{f}_2△t-(f_1△t-10g△t)\mu =9v^{\prime }27-40\times 0.4=9v^{\prime }\Rightarrow △t\to 0,10g△t\to 027-16=9v^{\prime }11=9v^{\prime }\Rightarrow v^{\prime }=\frac{11}{9}m/s$