Answer:
the ball experiences the greater momentum change
Explanation:
You have to take into account that momentum change is given by
[tex]\Delta p=mv_f-mv_b[/tex]
where vf and vb are the speed of the object after and before the impact.
In the case of the ball you have
[tex]\Delta p=(0.05kg)(-10.0\frac{m}{s})-(0.05kg)(10.0\frac{m}{s})=-1kg\frac{m}{s}[/tex]
where the minus of vf is included due to the motion is in an opposite direction regarding with vb
And for the lump
[tex]\Delta p=(0.05kg)(10.0\frac{m}{s})-(0.05kg)(0\frac{m}{s})=0.5kg\frac{m}{s}[/tex]
Hence, the ball experiences the greater change
hope this helps!!
