Question

Vic beasley (m = 102 kg) sacked braxton miller (m = 98 kg) a few times during the discover orange bowl. suppose that vic is chasing braxton at 0.25 m/s while braxton, trying to avoid the sack and gain positive yards, runs forward at -0.2 m/s. after the collision, braxton’s final velocity is 0 m/s. what is vic’s final velocity (m/s)? you may assume that the two are running in a straight line towards each other.

Answer 1

Using the conservation of momentum, Vic's final velocity is calculated to be 0.0578 m/s after he sacks Braxton, who comes to a stop.

Step-by-step explanation:

To solve this problem, we apply the principle of conservation of momentum. In a closed system (like the one described in the football scenario), the total momentum before collision is equal to the total momentum after the collision. The formula is given by:

m1 * v1initial + m2 * v2initial = (m1 + m2) * vfinal

Here, m1 is the mass of Vic Beasley (102 kg), m2 is the mass of Braxton Miller (98 kg), v1initial is Vic's initial velocity (0.25 m/s), and v2initial is Braxton's initial velocity (-0.2 m/s). The problem tells us that after the collision, Braxton's final velocity (v2final) is 0 m/s. Since they collided, we assume they do not cling to each other and thus calculate Vic's final velocity (v1final) separately:

102 kg * 0.25 m/s + 98 kg * (-0.2 m/s) = 102 kg * v1final + 98 kg * 0 m/s

25.5 kg*m/s - 19.6 kg*m/s = 102 kg * v1final

5.9 kg*m/s = 102 kg * v1final

v1final = 5.9 kg*m/s / 102 kg

v1final = 0.0578 m/s

Therefore, Vic's final velocity after the collision is 0.0578 m/s in the direction he was originally chasing Braxton.