Question

The blocks shown in figure (9-E19) have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. Take $g=10m/s^2.$

Answer 1

$V_1=10m/s$

$V_2=0,$

$V_1,V_2\to$ velocity of A and B after colliosion, (a) If the collision is perfectly elastic,

$m{V}_1+m{V}_2=M{v}_1+m{v}_2$

$\Rightarrow 10+0=v_1+v_2$

$v_1+v_2=10...\left(i\right)$

Again $v_1+v_2=-(u_1-v_2)$

$=-\left(10.0\right)=-10...\left(ii\right)$

Substrating (ii) from (i)

$2v_2=20$

$\Rightarrow v_2=10m/s$

The deceleration of B,

Putting work energy principle,

$\therefore \left(\frac{1}{2}\right)\times m\times (0{)}^2-\left(\frac{1}{2}\right)\times m\times v^2=-m\times a\times h$

$\Rightarrow -\left(\frac{1}{2}\right)\times (10{)}^2=-\mu g\times h$

$\Rightarrow h=\frac{100}{2\times 1\times 10}=50m$

(b) If the collision is perfectly in elastic.

$m\times u_1+m\times u_2=(m+m)\times v$

$\Rightarrow m\times 10+m\times 0=2m\times v$

$\Rightarrow v=\left(\frac{10}{2}\right)=5m/s$

The two blocks will move together sticking to each other.

$\therefore$ Putting work energy principle.

$\left(\frac{1}{2}\right)\times 2m\times (0{)}^2-\left(\frac{1}{2}\right)\times 2m\times (v{)}^2=2m\times \mu g\times S$

$\Rightarrow \frac{(5{)}^2}{0.1\times 0.1\times 2}=S$

$\Rightarrow S=12.5m.$