被弾性衝突

質問

質量m1の物体が速度v1で運動して、質量m2の静止している物体と衝突し、衝突後はv’1という速度になった。
この場合の反発係数eをv1、v’1、v’2で表せ。
また、運動量保存則を与え、衝突後のm2の速度v’2をm1、m2、e、v1を用いて表せ。
よろしくお願いします。

回答No.1

drmuraberg

Suushiki mo gaikoku-go no yohna monodesuna, maigosan.

The coefficient of restitution e is given by
e = (v’2-v’1)/(v1-v2)
for two colliding objects,

where
v’1 is the final velocity of the first object after impact,
v’2 is the final velocity of the second object after impact,
v1 is the initial velocity of the first object before impact,
v2 is the initial velocity of the second object before impact.

v’1 and v’2 can be written as
v’1= ｛m1v1+m2v2+m2e(v2-v1)｝/(m1+m2)
v’2= ｛m1v1+m2v2+m1e(v1-v2)｝/(m1+m2)

These equations can be derived from the analytical solution
to the system of equations formed by the definition of the e
and the law of the conservation of momentum (which holds for all
collisions).

m1v1 + m2v2 = m1v'1 +m2v'2
and
e = (v’2-v’1)/(v1-v2)

If you simply put v2=0, you will get the answers.

「被」 must be replaced by 「非」。