Question

Particle A has half the mass and twice the speed of particle B. Is the momentum p_A less than, greater than, or equal to p_B? Explain.

Short answer

P_{A $>$}P_B .

Thus momentum of particle 'A' is greater than momentum of particle 'B'.

Step-by-step solution

Step 1.  Given Information,

We have given,

M_A= _$1/2$M_B ,

V_A = 2V_B ,

 Step 2; Simplfy ,

^P_B $=$ ^P_B ^M_B ^V_B

$=$$\frac{M_{B_{}}{V}_B}{\sqrt{1-{\displaystyle \frac{{V_{b_{}}}^2}{c^2}}}}$

$P_A=P_A{M}_B{V}_A=\frac{M_A{V}_B}{\sqrt{1-{\displaystyle \frac{{V_A}^2}{c^2}}}}=\frac{{\displaystyle \frac{1}{2}}{M}_B\times 2V_B}{\sqrt{1-{\displaystyle \frac{4{V_B}^2}{c^2}}}}>P_B$

So, P_A > P_B

Thus momentum of particle A is greater than the the momentum of particle B .