The initial kinetic energy of the bullet is:
\[ K_i = \frac{1}{2} m (100)^2 \]
The final kinetic energy of the bullet is:
\[ K_f = \frac{1}{2} m (40)^2 \]
The percentage loss in kinetic energy is given by:
\[ \% \text{loss} = \frac{|K_f - K_i|}{K_i} \times 100 \]
Substituting the expressions for \( K_i \) and \( K_f \):
\[ \% \text{loss} = \frac{\left| \frac{1}{2} m (40)^2 - \frac{1}{2} m (100)^2 \right|}{\frac{1}{2} m (100)^2} \times 100 \]
Simplify:
\[ \% \text{loss} = \frac{|1600 - 10000|}{10000} \times 100 \]
\[ \% \text{loss} = \frac{8400}{10000} \times 100 = 84\% \]
Final Answer: 84% (Option 4)
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
Choose the correct set of reagents for the following conversion:
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):