Question

The K.E. of a body decreases by 19% what
is the percentage decrease in momentum
20 %
0 15 %
O
10 %
0 5 %​

Answer 1

Explanation:

K.E = P²/2m

K.E1/K.E2 = (P1/P2)²

K.E1 = 100%

SO, K.E2 = 100-19 = 81%

K.E2 = 1/2(mv²) × 19/100

K.E = K.E1 - K.E1× 19/100

K.E = 81 K.E1/100

MV² = 81mv1²/100

V² = 81/100(v1)²

V = 9/10(v1)

loss percent = total - loss /total × 100

= K.E1 - 9/10K.E1 /K.E1 ×100

= K.E1(1-9/10)/K.E1 ×100

= 1/10× 100

P = 10%

therefore , there is 10% decrease in momentum

Answer 2

Solution:-

we know that

$\sf \to \: K. E = \dfrac{1}{2} mv {}^{2}$

Now multiply and divide by M because we get formula of momentum

$\sf \to \: K. E = \dfrac{1}{2m} {m}^{2} {v}^{2} = \dfrac{1}{2m} (mv) {}^{2}$

$\sf \to \: K. E = \dfrac{1}{2m} \times {p}^{2} = \dfrac{ {p}^{2} }{2m}$

Now kinetic energy decreased by 19%

$\rm \to \: K. E_f = K. E_i - K. E_i \bigg( \dfrac{19}{100} \bigg)$

$\rm \to \: K. E_f = \dfrac{81}{100} K. E_i$

$\rm \to \: \dfrac{K. E_f}{K. E_i} = \dfrac{81}{100}$

Now we can write as

$\rm \to \: \dfrac{ \dfrac{p ^{2} _f}{2m} }{ \dfrac{ {p}^{2}_i }{2m} } = \dfrac{81}{100}$

$\rm \to \: \dfrac{ \dfrac{p ^{2} _f}{ \cancel{2m}} }{ \dfrac{ {p}^{2}_i }{ \cancel{2m}} } = \dfrac{81}{100}$

$\rm \to \: \bigg(\dfrac{p_f}{p_i} \bigg) ^{2} = \bigg( \dfrac{9}{10} \bigg)^{2}$

$\rm \to \: p_f = \dfrac{9}{10} p_i$

Formula % decrease Linear Momentum

$\rm \to \: \dfrac{p_f - p_i}{p_i} \times 100$

Put the value

$\rm \to \: \dfrac{\dfrac{9}{10} p_i - p_i}{p_i} \times 100$

$\rm \to \: \dfrac{p_i \bigg( \dfrac{9}{10} - 1 \bigg)}{p_i} \times 100$

$\rm \to \: \dfrac{9}{10} - 1 \times 100$

$\rm \to \: \dfrac{ - 1}{10} \times 100$

$\rm \to \: - 10 \%$

=> 10% Decrease