A neutron of mass 1.67 times 10^{-27} kg mov | vedclass

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Question

(C) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.Let $m_1 = 1

Vedclass · Accepted Answer

$3.33 	imes 10^7 \ m/s$

Vedclass · Answer

(C) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.Let $m_1 = 1.67 	imes 10^{-27} \ kg$ be the mass of the neutron and $v_1 = 10^8 \ m/s$ be its velocity.Let $m_2 = 3.34 	imes 10^{-27} \ kg$ be the mass of the deuteron,which is initially at rest $(v_2 = 0)$.Let $v$ be the final velocity of the combined system of mass $(m_1 + m_2)$.Applying the conservation of momentum:$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v$$(1.67 	imes 10^{-27}) 	imes 10^8 + (3.34 	imes 10^{-27}) 	imes 0 = (1.67 	imes 10^{-27} + 3.34 	imes 10^{-27})v$$1.67 	imes 10^{-19} = (5.01 	imes 10^{-27})v$$v = \frac{1.67 	imes 10^{-19}}{5.01 	imes 10^{-27}}$$v = 0.333 	imes 10^8 \ m/s = 3.33 	imes 10^7 \ m/s$.

$A$ neutron of mass $1.67 \times 10^{-27} \ kg$ moving with a velocity of $10^8 \ m/s$ collides with a deuteron at rest and sticks to it. If the mass of the deuteron is $3.34 \times 10^{-27} \ kg$,what will be the velocity of the combined system?

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