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lerasteidl
03.12.2020 •
Physics
A block is resting on a ramp the magnitude of the force of friction on the block is equal to
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Ответ:
4m/s
Explanation:
Mass of girl = m1 = 30kg
Mass of boy = m2 = 25kg
Resultant speed after collision = 1.0m/s
Speed of girl = v1
Speed of boy = v2
Momentum = mass × velocity
m1v1 - m2v2 = mv
30v1 - 25v2 = 1×m
30v1 - 25v2 = m
v1 = (m + 25v2)/30 ...eqn 1
but m1 - m2 = m
m = 30 - 25 = 5kg
v1 = (5 + 25v2)/30 ...eqn 2
v1-v2 = 1m/s
v2 = (v1 - 1)m/s
Substitute for v2 in eqn 2
v1 = [5 + 25(v1 - 1)]/30
30v1 = 5 + 25v1 - 25
30v1 - 25v1 = -20
5v1 = -20
v1 = -4m/s
Because she moved in an opposite direction to the boy, the negative sign is featured on her speed