A heavy truck and bike are moving with the same kinetic energy . If the mass of the truck is four times that of the bike , then calculate the ratio of their momenta . (Ratio of momenta = 1:2)

Explanation

(i) Given: Let `m_(b),m_(1)` are the masses of truck and bike,
`m_(1)=4m_(b)` . . . (1)
Here kinetic energies of both truck and bike are same
`(1)/(2)m_(1)v_(1)^(2)=(1)/(2)m_(b)v_(b)^(2)` . . (2)
`(1)/(2)xx4m_(b)vt^(2)=(1)/(2)m_b v_(b)^(2)`
`v_(t)^(2)=v_(b)^2`
`v_(t)^(2)=(1)/(4)v_(b)^(2)`
`v_(t)=(1)/(2)v_(b)`
To find ratio of the momenta, (i.e)
`m_(1):v_(1):m_(b)v_(b)`
`(P_(truck))/(P_(bike))=(m_(t)v_(t))/(m_(b)v_(b))`
`=4xx(1)/(2)=(2)/(1)=2:1`
(or) `(P_(bike))/(P_(truck))=(1)/(2)=1:2`
(ii) `v_(T)=(v_(o)+0.16T)`
`T=46^(@)C,v_(o)=331ms^(-1)`
`v_(T)=331+(0.61xx46)`
`=331+28.06`
`v_(T)=359.06ms^(-1)`.
`v_(T)=(v_(o)+0.16T)`
`T=46^(@)C,v_(o)=331ms^(-1)`
`v_(T)=331+(0.61xx46)`
`=331+28.06`
`v_(T)=359.06ms^(-1)`