Question

A ball of mass $m=1$ kg strikes smooth horizontal floor as shown in figure. The impulse exerted on the floor is

• A. $6.25Ns$
• B. $1.76Ns$
• C. $7.8Ns$
• D. $2.2Ns$

Answer 1

The correct option is A $6.25Ns$

Resolving the velocity before collision u into horizontal and vertical components we get

$u_x=5\cos {53}^0=5\ast \frac{3}{5}=3m/s$;

$u_y=5\sin {53}^0=5\ast \frac{4}{5}=4m/s$;

Resolving the velocity before collision v into horizontal and vertical components we get

$v_x=v\cos {37}^0=\frac{4}{5}v$;

$v_y=v\sin {37}^0=\frac{3}{5}v$;

Since component parallel to the plane remains unchanged, $u_x=v_x$;

$\Rightarrow 3=\frac{4}{5}v$;

$\Rightarrow v=\frac{15}{4}=3.75m/s$;

$\Rightarrow v_y=\frac{15}{4}\ast \frac{3}{5}=2.25m/s$;

$v_y$ is opposite to $u_x$;
$m=1kg$;
Hence, change in momentum perpendicular to the plane is $\left|\Delta p\right|=1\ast (4+2.25)=6.25kgNs$.