Answer:
0.62 m/s² at 68° S of E
Explanation:
Net force north = 12 - 70 = -58 N
Net force east = 33 - 10 = 23 N
Net force = √(-58² + 23²) = 62.3939... N
acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²
θ = arctan(-58/23) = -68.3691... ≈ 68° S of E
i think the data is not complete but that's according to me
Answer:
0.167m/s
Explanation:
According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.
Given momentum = Maas × velocity.
Momentum of glider A = 1kg×1m/s
Momentum of glider = 1kgm/s
Momentum of glider B = 5kg × 0m/s
The initial velocity of glider B is zero since it is at rest.
Momentum of glider B = 0kgm/s
Momentum of the bodies after collision = (mA+mB)v where;
mA and mB are the masses of the gliders
v is their common velocity after collision.
Momentum = (1+5)v
Momentum after collision = 6v
According to the law of conservation of momentum;
1kgm/s + 0kgm/s = 6v
1 =6v
V =1/6m/s
Their speed after collision will be 0.167m/s
