Question

A cyclist of mass 60.0 kg sits on a bicycle of mass 10.0 kg, moving at 15 m/s in the positive -direction. a bug of mass 5.0 g splatters on her cycle helmet. the bug was initially moving at 2.0 m/s in the same direction as the cyclist. a) what is the combined momentum of cycle and cyclist before colliding with the bug? b) what is the initial momentum of the bug? c) what is the change in the bicyclist’s velocity due to colliding with the bug? d) if the bug had been moving in the opposite direction to the cyclist, how much of a velocity change would have resulted from the collision?

Answer 1

Answer:

Part a)

Momentum of cyclist and cycle is 1050 kg m/s

Part b)

Momentum of bug is 0.05 kg m/s

Part c)

There is no change in the speed of cycle due to collision

Part d)

There is no change in the speed of cycle due to motion of bug

Explanation:

Part a)

Combined momentum of cycle and cyclist is given as

$P = (m_1 + m_2)v$

$P = (60 + 10)15$

$P = 1050 kg m/s$

Part b)

Initial momentum of the bug is given as

$P = 0.005 (2)$

$P = 0.01 kg m/s$

Part c)

Since the momentum of bug is very small as compare to the momentum of cycle + boy so there is no change in the momentum of cycle due to collision with the bug

Part d)

Since the bug is moving with very small momentum as compare to bicycle so there is no change in the momentum of cycle due to collision with the bug.

So there is no change in the speed of cycle

#Learn

Topic : momentum

https://brainly.in/question/5298071