Question

2. A heavy truck and bike are moving with<br />the same kinetic energy. If the mass of<br />the truck is four times that of the bike,<br />then calculate the ratio of their momenta,​

Answer 1

Answer:

p= mv

p is momentum

velocity is same.

let mass of bike be x

truck's mass = 4x

for truck

P1 = m1v

= 4xv

for bike

P2 = m2v

= xv

ratio = P1/P2

= 4xv/xv

= 4:1

Answer 2

Given. :-

Mass of truck = 4Mass of bike

•m1= m

•m2= 4m

• We have to find. the ratio of momentum of bike and truck

Then according to the kinetic energy formula

$\bigstar{\boxed{\sf{ K.E= \dfrac{1}{2}mv^2}}}$

where ,

K.E= Kinetic energy

m= Mass

v= velocity

Solution

$\longrightarrow\sf \dfrac{1}{2}m_1{(v_1)}^2= \dfrac{1}{2}m_2{(v_2)}^2$

• First find their velocity

$\longrightarrow\sf \bullet m_1= m\ \ \ \bullet m_2= 4m \\ \\ \longrightarrow\sf \cancel{\dfrac{1}{2}}m(v_1)^2= \cancel{\dfrac{1}{2}}\times 4m(v_2)^2\\ \\ \\ \longrightarrow\sf \cancel{m}v_1^2= 4\cancel{m}v_2^2\\ \\ \\ \longrightarrow\sf v_1^2= 4(v_2)^2\\ \\ \\ \longrightarrow\sf (v_1)^{\cancel2}= (2v_2)^{\cancel2}\\ \\ \\ \longrightarrow\sf v_1= 2v_2$

Now the ratio of their momentum

$\sf{\dag\ \ Ratio = \dfrac{Momentum\ of \ bike }{Momentum \ of \ truck }}$

$\longrightarrow\sf Ratio = \dfrac{m_1v_1}{m_2v_2}\\ \\ \\ \longrightarrow\sf Ratio = \dfrac{m\times 2\cancel{v_2}}{4m\times \cancel{v_2}}\\ \\ \\ \longrightarrow\sf Ratio = \cancel{\dfrac{2m}{4m}}\\ \\ \\ \longrightarrow\sf Ratio = \dfrac{1}{2}= 1:2$

$\underline{\boxed{\sf{ 1:2}}}$