1 Answer

Alioguzhan · Answer 1 · 2023-09-30T01:24:24+0000

Answer:

0.15 m/s

Step-by-step explanation;

The collision is inelastic, which means only momentum is conserved.

Now, before the collision the momentum of the system is

$p_{in\text{ital}}=(0.19\operatorname{kg})v+(0.85\operatorname{kg})\cdot0.70_{}(m/s)_{}$

After the collision the momentum becomes

$p_{\text{fInal}}=(0.19\operatorname{kg}+0.85\operatorname{kg})\cdot4.2$
$=0.728(kg\cdot m/s)$

The conservation of momentum demands that

$p_{\text{Iniital}}=p_{\text{fInal}}$

therefore, we have

$(0.19\operatorname{kg})v+(0.85\operatorname{kg})\cdot0.70(m/s)_{}=0.728(kg\cdot m/s$

Neglecting the units for a while gives us

subtracting 0.70 from both sides gives

finally, dividing both sides by 0.19 gives and rounding the nearest hundredth gives

$\boxed{v=0.15.}$

Hence, the velocity of the hockey puck before the collision is 0.15 m/s.

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