If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

Question

If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be : (A) 100 % (B) 200 % (C) 300 % (D) 400 %

Tardigrade Admin · Accepted Answer

K2=4 K1 (1/2) m v22=4 (1/2) m v12 v2=2 v1 P=m v P2=m v2=2 m v1 P1=m v1 % change =(Δ P/P1) × 100 =(2 m v1-m v1/m v1) × 100=100 %

Q. If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

$100%$

$200%$

$300%$

$400%$

$K_{2} = 4 K_{1}$
$\frac{1}{2} m v_{2}^{2} = 4 \frac{1}{2} m v_{1}^{2}$
$v_{2} = 2 v_{1}$
$P = m v$
$P_{2} = m v_{2} = 2 m v_{1}$
$P_{1} = m v_{1}$
$%$ change $= \frac{Δ P}{P _{1}} \times 100$
$= \frac{2 m v _{1} - m v _{1}}{m v _{1}} \times 100 = 100%$

Q. If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

100%

200%

300%

400%

K2​=4K1​ 21​mv22​=421​mv12​ v2​=2v1​ P=mv P2​=mv2​=2mv1​ P1​=mv1​ % change =P1​ΔP​×100 =mv1​2mv1​−mv1​​×100=100%

$100%$

$200%$

$300%$

$400%$

$K_{2} = 4 K_{1}$
$\frac{1}{2} m v_{2}^{2} = 4 \frac{1}{2} m v_{1}^{2}$
$v_{2} = 2 v_{1}$
$P = m v$
$P_{2} = m v_{2} = 2 m v_{1}$
$P_{1} = m v_{1}$
$%$ change $= \frac{Δ P}{P _{1}} \times 100$
$= \frac{2 m v _{1} - m v _{1}}{m v _{1}} \times 100 = 100%$