A 0.450-kg hockey puck, moving east with a speed of 5.80 m/s, has a head-on collision with a 0.900-kg puck initially at rest. As
1 answer:
Answer:
Explanation:
mass of first puck, m1 = 0.450 kg
initial velocity of first puck, u1 = 5.80 m/s east
mass of second puck, m2 = 0.9 kg
initial velocity of second puck, u2 = 0
Let v1 and v2 are the final velocities of the pucks after collision.
Use conservation of momentum
m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2
0.45 x 5.8 + 0.9 x 0 = 0.45 x v1 + 0.9 x v2
5.8 = v1 + 2 v2 ..... (1)
As the collision is perfectly elastic, the coefficient of restitution, e = 1
Use the formula for the coefficient of restitution
0 - 5.8 = v1 - v2
v2 - v1 = 5.8 .... (2)
Adding equation (1) and equation (2), we get
v2 = 1.93 m/s and then v1 = - 3.87 m/s
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Answer:
46.48m/s
Explanation:
The problem is a combination of the principle of conservation of linear momentum and projectile motion.
The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.
For two bodies of masses and
moving with velocities
and
before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);
were v is their common velocity after impact. If the second mass was at rest before the impact, then its initial velocity
. therefore
. Equation (1) then becomes;
In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.
Given;
Substituting these values into (2), we get the following;
Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;
were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;
were g is acceleration due to gravity taken as . Therefore;
We then substitute R and t into equation (4) to obtain v.
We now further substitute this value of v into (3) to obtain ;