Question

An ice hockey puck with a mass of 0.19 kg collides inelastically with a 0.85 kg snowball that is sliding to the left with a speed of 0.70 m/s. The combined puck and snowball slide along the ice with a velocity of 4.2 m/s to the right. What is the velocity of the hockey puck before the collision?

Answer 1

Answer:

0.15 m/s

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

$0.19v+0.70=0.728$

subtracting 0.70 from both sides gives

$0.19v=0.728-0.70$$0.19v=0.028$

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.