F = m.a
F = 3(2 I + 5 j)
F = 6i + 15j is resultant force
Magnitude force = √6+15 = √21
Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

Thus, the speed of the bullet-block center of mass is 5.52 m/s.
Answer
given,
initial speed,u = 0 m/s
final speed,v = 24.8 m/s
time, t= 5.9 s
a) acceleration,


a = 4.20 m/s²
b) distance to travel in that time
using equation of motion
v² = u² + 2 as
24.8² = 0² + 2 x 4.20 x s
8.4 s = 615.04
s = 73.22 m
the motorcycle will travel 73.22 m