Question

Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.500 kg. The two blocks have a perfectly inelastic collision and stick together after the collision. 0.200 m/s A y x B 0.400 m/s (a) What is the speed of the combined blocks after the collision? m/s (b) What angle does the velocity of the combined blocks make with the +x-axis after the collision? ° counterclockwise from the +x-axis (c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision? J Additional Materials Elastic Collisions

          Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.500 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.
0.200 m/s
A
y
x
B
0.400 m/s
(a) What is the speed of the combined blocks after the collision?
m/s
(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?
° counterclockwise from the +x-axis
(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?
J
Additional Materials
Elastic Collisions

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Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.500 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.
0.200 m/s
A
y
x
B
0.400 m/s
(a) What is the speed of the combined blocks after the collision?
m/s
(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?
° counterclockwise from the +x-axis
(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?
J
Additional Materials
Elastic Collisions

Added by Javier E.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Cyra Jelle Calleja

10/07/2021

Step 1

First, we need to find the initial momentum of the system. Momentum is a vector quantity, so we need to consider both the magnitude and direction of the velocities. For block A, the momentum is (0.500 kg)(0.200 m/s) = 0.100 kg*m/s in the +x direction. Show more…

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Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.500 kg. The two blocks have a perfectly inelastic collision and stick together after the collision. (a) What is the speed of the combined blocks after the collision? (b) What angle does the velocity of the combined blocks make with the +x-axis after the collision? counterclockwise from the +x-axis (c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision? Additional Materials Elastic Collisions

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00:01 In this problem, we have two blocks of the same mass of 0 .5 kilogram that is moving.

00:11 For block a, it is moving towards positive x direction with initial velocity of 0 .2 meter per second, while block b is moving initially at a speed of 0 .4 meter per second to positive y direction.

00:54 Now, question a want us to determine the final velocity of the two blocks if they stick together when they collide.

01:09 So to solve this problem, we will use the conservation of momentum.

01:13 This problem is an elastic collision.

01:16 So the momentum is conserved.

01:19 The momentum, we know that is the product of the mass and the velocity.

01:26 So if the momentum is conserved, the total momentum initially, is equal to the total momentum final.

01:33 So we have first for the block a, this is the initial momentum.

01:37 For block b, this is the initial momentum.

01:40 Then it's equals to final velocity multiplied by the sum of the masses since they move at that same velocity now, since they stay together.

01:55 And take note that the velocity here is a vector.

01:59 So the magnitude of the final velocity is equal to the square root of the sum of the of the squared of its vertical and horizontal components.

02:08 So the horizontal is x, x of x, and for the horizontal it's vf y.

02:17 So first we must determine for the horizontal component...

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