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At what speed relative to the lab will a 0.641-kg object have the same momentum as a 1.30-kg object that is moving at 0.515c relative to the lab?

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Borys Shumyatskiy eNotes educator | Certified Educator

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Hello!

Denote the given masses as `m_1^0` and `m_2^0,` and the speeds as `v_1` and `v_2.` I suppose the given masses are the rest masses, and they'll change with the changing of the speeds.

Relativistic momentum is defined the same way as non-relativistic momentum: `p=mv,` where `m` is the current (observed) mass depending of a frame of reference. Also we know that the mass `m` depends on the speed `v` as `m=m_0/sqrt(1-v^2/c^2),` so

`p_1 = m_1 v_1 = (m_1^0 v_1)/sqrt(1-v_1^2/c^2),`   `p_2 = m_2 v_2 = (m_2^0 v_2)/sqrt(1-v_2^2/c^2).`

It is given that `p_1=p_2.` The only unknown quantity here is `v_1,` and we can find it from this equation. It is more convenient to express `v_1` in terms of `c,` i.e. to find `k_1=v_1/c` (`k_2=v_2/c` is given). The equation is then equivalent to

`(m_1^0 k_1)/sqrt(1-k_1^2) = (m_2^0 k_2)/sqrt(1-k_2^2),`  or squared  `((m_1^0)^2 k_1^2)/(1-k_1^2) = ((m_2^0)^2 k_2^2)/(1-k_2^2).`

Multiply by `1-k_1^2,` collect the terms with `k_1^2` and it becomes

`k_1^2((m_1^0)^2+((m_2^0)^2 k_2^2)/(1-k_2^2)) = ((m_2^0)^2 k_2^2)/(1-k_2^2),`

and finally

`k_1^2 = (((m_2^0)^2 k_2^2)/(1-k_2^2)) /((m_1^0)^2+((m_2^0)^2 k_2^2)/(1-k_2^2)).`

To find the number, first compute `((m_2^0)^2 k_2^2)/(1-k_2^2) = (1.30)^2*(0.515)^2/(1-(0.515)^2) approx 0.61.` Then `k_1^2 approx 0.61/((0.641)^2+0.61) approx 0.60.` And `k_1 approx sqrt(0.60) approx 0.77.` So the answer is: the first object must have a speed of about 0.77c.

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