Question

e. which, therefore, suffers the greater damage? (bus) (both the same) (the bug of course!)
the bug of course!

1. granny whizzes around the rink and is suddenly confronted with ambrose at rest directly in her

path. rather than knock him over, she picks him up and continues in motion without \braking.\
consider both granny and ambrose as two parts of one system. since no outside forces act on
the system, the momentum of the system before collision equals the momentum of the system
after collision.

a. complete the before - collision data in the table below.
before collision
granny’s mass 80 kg
granny’s speed 3 m/s
granny’s momentum ______
ambrose’s mass 40 kg
ambrose’s speed 0 m/s
ambrose’s momentum ______
total momentum ______

b. after the collision, does granny’s speed increase or decrease?

c. after collision, does ambrose’s speed increase or decrease?

d. after collision, what is the total mass of granny + ambrose?
120kg 80 + 40 = 120kg

e. after the collision, what is the total momentum of granny + ambrose?

f. use the conservation of momentum law to find the speed of granny and
ambrose together after collision. (show your work in the space below.)

Explanation:

Step1: Calculate Granny's momentum

Momentum formula: \(p = mv\).
\(p_{Granny}=80\times3 = 240\space kg\cdot m/s\)

Step2: Calculate Ambrose's momentum

\(p = mv\), \(v = 0\space m/s\)
\(p_{Ambrose}=40\times0=0\space kg\cdot m/s\)

Step3: Calculate total momentum

\(p_{total}=p_{Granny}+p_{Ambrose}\)
\(p_{total}=240 + 0=240\space kg\cdot m/s\)

Step4: Answer part b

Since total momentum is conserved (\(p = mv\)), mass increases (\(m_{new}=m_{Granny}+m_{Ambrose}\)), so \(v\) (Granny's speed) decreases.

Step5: Answer part c

Ambrose was at rest (\(v = 0\)) before, after collision he moves, so his speed increases.

Step6: Answer part d

\(m_{total}=80 + 40=120\space kg\)

Step7: Answer part e

By conservation of momentum (no external forces), \(p_{total\space after}=p_{total\space before}=240\space kg\cdot m/s\)

Step8: Answer part f

Using \(p = mv\), \(v=\frac{p}{m}\)
\(v=\frac{240}{120}=2\space m/s\)

Answer:

a. Granny's momentum: \(240\space kg\cdot m/s\), Ambrose's momentum: \(0\space kg\cdot m/s\), Total momentum: \(240\space kg\cdot m/s\)
b. Decrease
c. Increase
d. \(120\space kg\)
e. \(240\space kg\cdot m/s\)
f. \(2\space m/s\)