Two campers dock a canoe. One camper has a mass of 100.0 kg and moves forward at 3.0 m/s as he leaves the canoe to step onto the
1 answer:
Answer:
The combined speed of camper and canoe is 1.71 m/s.
Explanation:
Given that,
Mass of camper 1, m = 100 kg
Speed of camper 1, v = 3 m/s
The combined mass of another camper and canoe is, M = 175 kg
We need to find the combined speed of camper and canoe. According to the conservation of linear momentum, the momentum of first camper is equal to linear momentum of the canoe and the second camper.
So, the combined speed of camper and canoe is 1.71 m/s.
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Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of