Question

A 39 kg girl runs across a mat with a speed of 4.8 m/s and jumps onto a 125 kg hanging platform initially at rest, causing the girl and platform to swing back and forth like a pendulum together after her jump. What is the combined velocity of the girl and platform (in m/s) after the jump?

Answer 1

Answer:

The combined velocity of the girl and the platform after the jump is 1.14 m/s.

Explanation:

From the law of conservation of momentum:

m1u1 = (m1 + m2)v

m1 is mass of the girl = 39 kg

m2 is mass of the hanging platform = 125 kg

u1 is initial speed of the girl = 4.8 m/s

v is combined velocity of the girl and the platform after the jump

v = m1u1/(m1+m2) = 39×4.8/(39+125) = 187.2/164 = 1.14 m/s

Answer 2

Answer:

The combined velocity of the platform and the girl is 1.14 m/s

Explanation:

This case can be simply treated as an inelastic collision between two objects.

However, in this case, it is the between the girl and the platform.

in an inelastic collision, the total momentum is conserved.

This means that:

Total momentum before collision = total momentum after collision.

momentum of running girl = momentum of swinging platform + girl

$mv(girl)= mv(platform+girl)$

$39\times 4.8=(125+39) \times v$

evaluating the above equation, we have

$187.2 = 164v$

$v = \frac{182.7}{164}$

$v=1.14$

$v= 1.14m/s$

∴ The combined velocity of the girl and the platform is 1.14 m/s