> #mathematics , #physics > When a 0.45g air rifle pallet is fired in target attache to glider on linear air track, the glider moves 50cm in 3.8s.Caluclate velocity of pellet.The glider and target have combine mass of 643g?
###### Nitha Mulungu
more than 3 weeks ago

# When a 0.45g air rifle pallet is fired in target attache to glider on linear air track, the glider moves 50cm in 3.8s.Caluclate velocity of pellet.The glider and target have combine mass of 643g?

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ChatGPT
more than 3 weeks ago
First, we need to calculate the momentum of the pellet before it hits the glider.

Given:
Mass of pellet (m1) = 0.45g = 0.00045kg
Initial velocity of pellet (u1) = 0 m/s (as it is fired from rest)

Momentum of pellet before hitting the glider:
p1 = m1 * u1 = 0.00045kg * 0 = 0 kg m/s

Now, we need to calculate the velocity of the glider after the pellet hits it.

Given:
Combined mass of glider and target (m2) = 643g = 0.643kg
Distance moved by glider (s) = 50cm = 0.5m
Time taken for glider to move (t) = 3.8s

Using the equation of motion:
s = (u2 + u1) * t / 2
0.5 = (u2 + 0) * 3.8 / 2
u2 = 0.5 * 2 / 3.8
u2 = 0.2632 m/s

Now, we can calculate the velocity of the pellet after hitting the glider using the principle of conservation of momentum:

Initial momentum = Final momentum
m1 * u1 = (m1 + m2) * u2
0 = (0.00045 + 0.643) * u2
u2 = 0 / 0.64345
u2 = 0 m/s

Therefore, the velocity of the pellet after hitting the glider is 0 m/s.
0
more than 3 weeks ago
Momentum of pellet and glider was conserved.
Momentum=mass x velocity
Momentum of pellet before firing=0 as it is at rest
Momentum of pellet after firing=0.45g(V)
Momentum of glider and target before firing=0(at rest)
Momentum of glider and target after firing=(643+0.45)g(u)
=643.45gu
Momentum of pellet=momentum of glider and target
0.45gV=643.45gu
V=(643.45gu)/0.45g
Velocity of pellet(V)=1430u

Therefore since u=50cm/3.8s
V=1430(50/3.8)
V=18842.1cm/s
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