1 answer

A 48.0 kg child is stationary in the middle of an ice rink (no friction)

Question:

A 48.0 kg child is stationary in the middle of an ice rink (no friction). A 0.750 kg football is thrown with a velocity of 21.2 m/s moving eastward. The child catches the ball and together they slide off in the eastward direction with a velocity of 0.326 m/s.
a.) Find the impulse (magnitude and direction) experienced by the football during the collision.
b.) Find the impulse (magnitude and direction) experienced by the child during the collision.
c.) Suppose the collision laster 0.254 sec. Find the average force (mag and direction) experienced by the child during the collision.
d.) What was the child's average acceleration during the collision?
e.) Characterize the collision as elastic, inelastic, or completely inelastic.
Please answer with complete instructions on how to derive the answer to the question. This is for a practice test that will be like my test, I just need to know how to do it in case I get something else like it.

Answers

a)Impulse experienced by football is \(m v=0.75 \times 21.2=15.9 \mathrm{kgm} / \mathrm{s}\)

Direction is eastwards.

b)Impulse experienced by child is \(m v=48 \times 0.326=15.63 \mathrm{kgm} / \mathrm{s}\)

Direction is eastwards.

c)Force experienced is \(\frac{m v}{\Delta t}=\frac{15.63}{0.254}=61.54 N\). Direction is eastwards

d)Average acceleration is \(\frac{v}{\Delta t}=\frac{0.326}{0.254}=1.28 \mathrm{~m} / \mathrm{s}^{2}\)

e)Initial momentum is \(m v=0.75 \times 21.2=15.9 \mathrm{kgm} / \mathrm{s}\)

Final momentum is \((m+M) v=(48+.75) \times .326=15.9 \mathrm{kgm} / \mathrm{s}\)

Initial kinetic energy is

$$ \frac{1}{2} m v^{2}=\frac{1}{2} \times 0.75 \times 21.2^{2}=168.54 J $$

Final Kinetic enegy

$$ \frac{1}{2}(m+M) v^{2}=\frac{1}{2} \times 48.75 \times 0.326^{2}=2.6 J $$

.

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