1 Calculate the man's speed just before he grabs the boy. Ignore the effects of friction. 6.2 Is the collision elastic? Use a calculation to support your answer.

Mass of man, $m_m = 87 \text{ kg}$
Mass of boy, $m_b = 22 \text{ kg}$
Initial velocity of boy, $v_{b,i} = 0 \text{ m/s}$ (standing still)
Final velocity of man and boy together, $v_f = 2.4 \text{ m/s}$

The total momentum before the collision equals the total momentum after the collision.

Rounding to two decimal places:

The man's speed just before he grabs the boy is $\boxed{\text{3.01 m/s}}$.

Using $v_{m,i} = 3.00689... \text{ m/s}$ from the previous calculation for accuracy:

Since $KE_i \neq KE_f$, kinetic energy is not conserved.

The collision is not elastic because the total kinetic energy before the collision (393.29 J) is not equal to the total kinetic energy after the collision (313.92 J). This is characteristic of an inelastic collision.

Combined mass, $M = m_m + m_b = 87 \text{ kg} + 22 \text{ kg} = 109 \text{ kg}$
Initial velocity on gravel, $v_i' = 2.4 \text{ m/s}$ (this is $v_f$ from the collision)
Final velocity on gravel, $v_f' = 0 \text{ m/s}$ (comes to rest)
Distance traveled on gravel, $d = 2 \text{ m}$

The negative sign indicates deceleration.

The magnitude of the force exerted by the gravel surface is $\boxed{\text{157.0 N}}$.